code forces 276C Little Girl and Maximum Sum （线段树/技巧）

The little girl loves the problems on array queries very much.

One day she came across a rather well-known problem: you've got an array of n elements (the elements of the array are indexed starting from 1); also, there are q queries, each one is defined by a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n). You need to find for each query the sum of elements of the array with indexes from l_i to r_i, inclusive.

The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 2·10⁵) and q (1 ≤ q ≤ 2·10⁵) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_i (1 ≤ a_i ≤ 2·10⁵) — the array elements.

Each of the following q lines contains two space-separated integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the i-th query.

Output

In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Example

Input

3 35 3 21 22 31 3

Output

25

Input

5 35 2 4 1 31 52 32 3

Output

33

【题解】

 刚开始用线段树做的，因为自己比较水，超时了，后来思考良久，发现了新大陆，没那么复杂，我们只需要把各点的查询频率排序，然后再把数排序，对应值相乘求和就好了。

【AC代码】

#include<cstdio>#include<algorithm>#include<cstdio>#include<math.h>#include<string.h>#include<iostream>using namespace std;typedef __int64 ll;const int N = 2e5+10;int a[N],c[N];int m,n;ll s;int main(){    while(~scanf("%d%d",&m,&n))    {        for(int i=1;i<=m;++i)            scanf("%d",&a[i]);        memset(c,0,sizeof c);        while(n--)        {            int x,y;            scanf("%d%d",&x,&y);            c[x]++;            c[y+1]--;        }        for(int i=1;i<=m;++i)            c[i]+=c[i-1];        sort(a+1,a+m+1);        sort(c+1,c+m+1);        s=0;        for(int i=1;i<=m;++i)            s+=ll(c[i])*ll(a[i]);        printf("%I64d\n",s);    }    return 0;}