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#include <algorithm>
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#include <cstdio>
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#include <vector>
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using namespace std;
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//快读
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int read() {
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int x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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const int N = 2e5 + 10;
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int n, m;
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int a[N];
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vector<int> ys;
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struct Node {
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int l, r, cnt;
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} tr[N << 5];
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int root[N], idx;
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int find(int x) {
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return lower_bound(ys.begin(), ys.end(), x) - ys.begin();
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}
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//经典的主席树插入
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void insert(int &u, int l, int r, int x) {
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tr[++idx] = tr[u]; //新开一个节点idx++,将新节点指向旧的tr[u]
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u = idx; //因为是引用，为了回传正确值，需要u=idx-1
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tr[u].cnt++; //新节点的cnt，因为多插入了一个数字，所以个数+1,这样处理的话，省去了pushup
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if (l == r) return; //如果已经到了叶子节点，上面的操作就足够了，可以直接返回，否则需要继续向下递归
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int mid = (l + r) >> 1;
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if (x <= mid)
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insert(tr[u].l, l, mid, x); //因为tr[u]进入本函数时，最先把旧的复制过来，所以tr[u].l也是上一个版本的左儿子节点
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else
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insert(tr[u].r, mid + 1, r, x);
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}
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int query(int p, int q, int l, int r, int k) {
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if (l == r) return l;
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int mid = (l + r) >> 1;
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int cnt = tr[tr[q].l].cnt - tr[tr[p].l].cnt;
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if (k <= cnt)
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return query(tr[p].l, tr[q].l, l, mid, k);
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else
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return query(tr[p].r, tr[q].r, mid + 1, r, k - cnt);
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}
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int main() {
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//文件输入输出
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#ifndef ONLINE_JUDGE
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freopen("P3834.in", "r", stdin);
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#endif
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n = read(), m = read();
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for (int i = 1; i <= n; i++) {
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a[i] = read();
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ys.push_back(a[i]);
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}
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//数据范围太大，直接建线段树会MLE，但是比较稀疏，可以离散化后用相对应的序号，数据量就小了
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sort(ys.begin(), ys.end());
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ys.erase(unique(ys.begin(), ys.end()), ys.end());
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// 0号版本没有内容时，主席树是不需要进行build的,强行build时，可能会有部分测试点TLE
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// 0号版本有内容时，主席树是需要build的，不build，初始值无法给上
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//主席树的数字增加，每增加一个，就相当于增加了一个版本root[i]记录了版本i的根节点
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for (int i = 1; i <= n; i++) {
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root[i] = root[i - 1];
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insert(root[i], 0, ys.size() - 1, find(a[i]));
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}
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while (m--) {
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int l, r, k;
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l = read(), r = read(), k = read();
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printf("%d\n", ys[query(root[l - 1], root[r], 0, ys.size() - 1, k)]);
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}
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return 0;
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}
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