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// 权值线段树解法
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 5e5 + 10;
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int a[N], b[N];
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int n;
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// 线段树区间和模板
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#define ls u << 1
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#define rs u << 1 | 1
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struct Node {
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int l, r; // 范围
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int sum; // 区间内数字个数
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} tr[N << 2]; // 4倍空间
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// 创建
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void build(int u, int l, int r) {
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tr[u].l = l, tr[u].r = r;
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if (l == r) return;
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int mid = l + r >> 1;
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build(ls, l, mid), build(rs, mid + 1, r);
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}
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// 向上更新统计信息
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void pushup(int u) {
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tr[u].sum = tr[ls].sum + tr[rs].sum;
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}
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// 修改点的值
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void modify(int u, int x, int v) {
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if (tr[u].l == x && tr[u].r == x) {
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tr[u].sum += v;
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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)
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modify(ls, x, v);
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else
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modify(rs, x, v);
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pushup(u);
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}
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// 查询区间和
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int query(int u, int l, int r) {
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if (l > r) return 0;
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if (tr[u].l == l && tr[u].r == r) return tr[u].sum;
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int mid = tr[u].l + tr[u].r >> 1;
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if (mid < l) return query(rs, l, r);
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if (r <= mid) return query(ls, l, r);
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return query(ls, l, mid) + query(rs, mid + 1, r);
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}
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LL ans;
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int main() {
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cin >> n;
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// 一般使用线段树，都喜欢从下标1开始，所以我们在处理数据时尽量让下标从1开始
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// b[]在下面的代码中有两个阶段的含义，目前的含义是a[i]的备份，因为一会a[]由小到大排序后，原始数组就丢失了位置信息就不见了，需要b备份出来
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for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
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// ① 原始数组由小到大排序
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sort(a + 1, a + n + 1);
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// ② 离散化，遍历原始数组b[i],找出它的排名，即b[2]=3表示第2个输入的是排名第3的
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for (int i = 1; i <= n; i++) b[i] = lower_bound(a + 1, a + n + 1, b[i]) - a;
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// 构建线段树
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build(1, 1, n); // n个数字
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// 边查边添加
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for (int i = 1; i <= n; i++) {
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modify(1, b[i], 1); // 占据b[i]这个位置
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ans += query(1, b[i] + 1, n); // 出现的比我早，号还比我大的有多少个
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}
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// 输出
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printf("%lld\n", ans);
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return 0;
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}
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