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#include <iostream>
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#include <unordered_map>
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#include <algorithm>
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#include <cstdio>
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#include <cstring>
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using namespace std;
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typedef long long LL;
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const int N = 30010, M = 100010;
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/*
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图灵树的来历原来是这个。。经典的线段树离线所有查询右端点排序的题目吧。。
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所有询问右端点排序后，从小到大扫过去，线段树维护序列区间和，用一个map记录各个数最右边出现的位置，
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一遇到一个数就把之前位置消除并更新当前位置，相当于把各个数尽量向右移动。。
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*/
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struct Node {
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int l, r;
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LL sum;
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} tr[N << 2];
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struct Question {
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int l, r, id;
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const bool operator<(const Question &t) const {
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return r < t.r;
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}
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} q[M];
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int n, m;
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int a[N];
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LL ans[M];
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void pushup(int u) {
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tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
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}
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void build(int u, int l, int r) {
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tr[u] = {l, r};
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int mid = (l + r) >> 1;
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if (l == r) return;
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build(u << 1, l, mid);
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build(u << 1 | 1, mid + 1, r);
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}
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void modify(int u, int x, int d) {
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if (tr[u].l == tr[u].r) {
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tr[u].sum += (LL)d;
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return;
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}
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int mid = (tr[u].l + tr[u].r) >> 1;
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if (x <= mid) modify(u << 1, x, d);
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if (x > mid) modify(u << 1 | 1, x, d);
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pushup(u);
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}
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LL query(int u, int l, int r) {
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if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum;
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LL ans = 0;
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int mid = (tr[u].l + tr[u].r) >> 1;
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if (l <= mid) ans += query(u << 1, l, r);
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if (r >= mid + 1) ans += query(u << 1 | 1, l, r);
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return ans;
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}
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int main() {
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// 加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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int T;
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cin >> T;
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while (T--) {
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unordered_map<int, int> b;
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memset(ans, 0, sizeof ans);
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i];
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cin >> m;
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for (int i = 1; i <= m; i++) {
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cin >> q[i].l >> q[i].r;
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q[i].id = i;
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}
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// 按右端点排序
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sort(q + 1, q + 1 + m);
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// 构建线段树
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build(1, 1, n);
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for (int i = 1, j = 1; i <= m; i++) { // i枚举每个查询,j枚举的是原始数组数据
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while (j <= q[i].r) { // 在空间内的逐个进入，并不断修改最后的值
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if (b.count(a[j])) modify(1, b[a[j]], -a[j]);
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modify(1, j, a[j]);
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b[a[j]] = j++;
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}
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ans[q[i].id] = query(1, q[i].l, q[i].r);
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}
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for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
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}
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return 0;
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}
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