#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 3e5 + 10;
int n;
int q[N];
int t[N];
LL ans;

/*
6
0 1 2 3 4 5
答案：15
*/

// 归并排序，附带求顺序对的数量
void merge_sort(int q[], int l, int r) {
  if (l >= r)
    return;
  int mid = (l + r) >> 1;
  merge_sort(q, l, mid), merge_sort(q, mid + 1, r);
  int i = l;
  int j = mid + 1;
  int k = 0;
  while (i <= mid && j <= r)
    if (q[i] <= q[j]) {
      t[k++] = q[i++];
      // 顺序对，原理：当q[i]<=q[j]时，q[i]与q[j]之后的所有数都构成顺序对
      ans += r - j + 1;
    } else
      t[k++] = q[j++];

  while (i <= mid)
    t[k++] = q[i++];
  while (j <= r)
    t[k++] = q[j++];
  for (i = l, j = 0; i <= r; i++, j++)
    q[i] = t[j];
}

int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> q[i];
  // 根据题目提示，使用归并排序，并在归并排序的过程中计算顺序对的数量
  merge_sort(q, 1, n);
  printf("%lld", ans);
  return 0;
}