/*
逆序对数量：
归并排序解法 https://blog.csdn.net/m0_50299150/article/details/122729602
树状数组解法 https://blog.csdn.net/m0_50299150/article/details/122729602
权值线段树解法
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 5e5 + 10;
int a[N];
vector<int> nums;
//手写二分版本+离散化 	325 ms
int n;
struct node {
    int l, r, sum;
} tr[N << 2];

void build(int root, int l, int r) {
    tr[root] = {l, r, 0};
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(root << 1, l, mid);
    build(root << 1 | 1, mid + 1, r);
}
void pushup(int root) {
    tr[root].sum = tr[root << 1].sum + tr[root << 1 | 1].sum;
}
void modify(int root, int pos, int val) {
    if (tr[root].l == pos && tr[root].r == pos) {
        tr[root].sum += val;
        return;
    }
    int mid = tr[root].l + tr[root].r >> 1;
    if (mid >= pos)
        modify(root << 1, pos, val);
    else
        modify(root << 1 | 1, pos, val);
    pushup(root);
}
int query(int root, int l, int r) {
    if (tr[root].l == l && tr[root].r == r) return tr[root].sum;
    int mid = tr[root].l + tr[root].r >> 1;
    if (mid < l)
        return query(root << 1 | 1, l, r);
    else {
        if (r <= mid)
            return query(root << 1, l, r);
        else
            return query(root << 1, l, mid) + query(root << 1 | 1, mid + 1, r);
    }
}
//改成手写二分查询，AC掉洛谷此题
int find(int x) {
    int l = 0, r = nums.size() - 1;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (nums[mid] >= x)
            r = mid;
        else
            l = mid + 1;
    }
    return l + 1;
}
int main() {
    //优化读入
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        nums.push_back(a[i]);
    }
    //离散化
    sort(nums.begin(), nums.end());
    nums.erase(unique(nums.begin(), nums.end()), nums.end());

    LL ans = 0;
    build(1, 1, nums.size() + 1);
    //边查边添加,查找它后面比它大的数字个数
    for (int i = 1; i <= n; i++) {
        int p = find(a[i]);
        ans += query(1, p + 1, nums.size() + 1);
        modify(1, p, 1);
    }
    cout << ans << endl;
    return 0;
}