#include<iostream>
using namespace std;

const int N = 1e6 + 10;
int num[N], tmp[N];
long long ans = 0;      //最大值n**2确实要考虑溢出的情况

//思路：归并排序里计数

void solve(int l, int r){
    //定义出口
    if(l == r){
        return ;
    }
    int mid = (l + r) / 2;
    //分治
    solve(l, mid);
    solve(mid + 1, r);


    //合并
    int i = l, j = mid + 1;
    int k = l;
    while(i <= mid && j <= r){
        //注意这边的等号
        if(num[i] <= num[j]){
            tmp[k++] = num[i++];
        }else{
            ans += (long long)(mid - i + 1);        //应该是和i相关，而不是j
            tmp[k++] = num[j++];
        }
    }
    while(i <= mid){
        tmp[k++] = num[i++];
    }
    while(j <= r){
        tmp[k++] = num[j++];
    }
    //复制数组
    for(int i = l; i <= r; i++)
        num[i] = tmp[i];

}

int main(){
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> num[i];

    solve(0, n - 1);

    cout << ans;

    return 0;
}