N - Little Girl and Maximum Sum——贪心

Description
   

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 %I64dspecifier.

Sample Input

3 35 3 21 22 31 3

Output

25

Input

5 35 2 4 1 31 52 32 3

Output

33

这道题的大意就是给你n个数，p次操作，

这p次操作输入l，r这个区间边界，求出区间内的值总和。但是这里有个条件是这n个数可以任意改变位置。

思路：这里我们定义一个cnt【】数组 ，表示l，r这两个点被累加了多少次，由于我们要用dp[i]=dp[i-1]+cnt[i]求出每个点被累加了多少次，所以为了防止当输入一个区间l，r时，不在区间里面的数累加，所以我们这里这样做cnt【l】++，cnt【r+1】--。

最后排个序，大的乘大的，小的乘小的，加起来就ok。

可能没有说清楚，哎，语文太差！！！！！

#include<string.h>
#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<math.h>
#define mx 200009
#define LL long long 
using namespace std;
LL     a[mx];
LL    dp[mx];
LL    cnt[mx];
bool cmp(LL    x,LL    y)
{
    return x>y;
}
int main()
{


    LL    l,r;
    LL    i,j;


    LL    n,q;
    while(scanf("%lld %lld",&n,&q)!=EOF)
    {
        memset(cnt,0,sizeof(cnt));
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
            scanf("%lld",&a[i]);
        while(q--)
        {
            scanf("%lld %lld",&l,&r);
            cnt[l]++;
            cnt[r+1]--;
        }
        for(i=1;i<=n;i++)
            dp[i]=dp[i-1]+cnt[i];
        LL    sum=0;
        sort(a,a+n,cmp);
        sort(dp,dp+n+1,cmp);


        for(i=0;i<n;i++)
       sum+=a[i]*dp[i];
        printf("%lld\n",sum);
    }
}

要用Long long，不然W