Stas and the Queue at the Buffet

During a break in the buffet of the scientific lyceum of the Kingdom of Kremland, there was formed a queue of nn high school students numbered from 11 to nn. Initially, each student ii is on position ii. Each student ii is characterized by two numbers — aiai and bibi. Dissatisfaction of the person ii equals the product of aiai by the number of people standing to the left of his position, add the product bibi by the number of people standing to the right of his position. Formally, the dissatisfaction of the student ii, which is on the position jj, equals ai⋅(j−1)+bi⋅(n−j)ai⋅(j−1)+bi⋅(n−j).

The director entrusted Stas with the task: rearrange the people in the queue so that minimize the total dissatisfaction.

Although Stas is able to solve such problems, this was not given to him. He turned for help to you.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of people in the queue.

Each of the following nn lines contains two integers aiai and bibi (1≤ai,bi≤1081≤ai,bi≤108) — the characteristic of the student ii, initially on the position ii.

Output

Output one integer — minimum total dissatisfaction which can be achieved by rearranging people in the queue.

Examples

Input

3
4 2
2 3
6 1

Output

12

Input

4
2 4
3 3
7 1
2 3

Output

25

Input

10
5 10
12 4
31 45
20 55
30 17
29 30
41 32
7 1
5 5
3 15

Output

1423
本题的题意大致为，n个同学，他有两个值，ai和bi，而他的不满意的程度是ai⋅(j−1)+bi⋅(n−j)ai⋅(j−1)+bi⋅(n−j).我们通过分解因式
得到（ai-bi）j+bi*n-ai
我们就可以通过一个数组吧ai-bi的值进行排序，大的结果乘以小的结果即可，接下来是代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long 
const int maxn = 200010;
ll a[maxn], b[maxn];
ll lazy[maxn];
int main()
{
    int t;
    cin >> t;
    ll sum = 0;
    ll ans = 0;
    for (int i = 0; i < t; i++)
    {
        scanf("%lld %lld", &a[i], &b[i]);
        sum += b[i] * t - a[i];
        lazy[i] = a[i] - b[i];
    }
    sort(lazy, lazy + t);
    for (int i = 1;i <= t;i++)
    {
        ans += lazy[t - i ] * i;
    }
    printf("%lld", ans + sum);

}