//小和问题，归并排序解决的问题的一种，将问题归纳为左半的值，加右边的值，加跨越左边和右边的值，以及这个问题能否排序解决
//https://www.nowcoder.com/practice/edfe05a1d45c4ea89101d936cac32469

#include <iostream>
#include <stdio.h>
using namespace std;

typedef long long LL;
const int N = 100010;
int arr[N] , help[N] , n;

LL marge(int left , int right , int mid){
    LL ans = 0;
    int temp = 0;
    int l = left , r = mid+1 , k = left;
    for(int i=left,j=mid+1;j<=right;j++){

        while(i <= mid && arr[i] <= arr[j]) temp+=arr[i++];
        ans += temp;
    }
    //cout<<left<<" "<<mid<<" "<<right<<endl;


    while(l <= mid && r <= right){
        if(arr[l] <= arr[r]){
            help[k++] = arr[l++];
        }else{
            help[k++] = arr[r++];
        }
    }

    while(l<=mid) help[k++] = arr[l++];
    while(r<=right) help[k++] = arr[r++];
    for(l=left;l<=right;l++){
        arr[l] = help[l];
    }
    return ans;
}

LL smallsum(int left , int right){
    if(left >= right) return 0;
    int mid = (left+right)/2;

    return smallsum(left,mid)+smallsum(mid+1,right)+marge(left,right,mid);
}


int main() {
    cin>>n;
    for(int i=1;i<=n;i++) cin>>arr[i];
    LL ans = smallsum(1,n);
    //for(int i=1;i<=n;i++) cout<<arr[i]<<" ";
    printf("%lld",ans);

}
// 64 位输出请用 printf("%lld")