CodeForces - 276C Little Girl and Maximum Sum 扫描线

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 nelements (the elements of the array are indexed starting from 1); also, there are qqueries, 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.

Examples

Input

3 3
5 3 2
1 2
2 3
1 3

Output

25

Input

5 3
5 2 4 1 3
1 5
2 3
2 3

Output

33

题意：给出n个数，q个查询，查询区间【u，v】每个点加1，求n个点的值与n个数的乘积和最大；

最先想到的是线段树的区间更新模板，然后觉得代码量有点大不想打，看了一下百度，觉得这个扫描线这个方法挺好的

#include<iostream>
#include<cstdio>
#include<string>
#include<map>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
const ll len=2e5+5;
ll arr[len];
ll num[len];
int main()
{
	int n,q;
	while(cin>>n>>q)
	{
		for(int i=0;i<n;++i)
			scanf("%lld",arr+i);
		int s,e;
		while(q--)
		{
			scanf("%d%d",&s,&e);
			num[s-1]++;num[e]--;
		}
		for(int i=1;i<n;++i)
			num[i]+=num[i-1];
		sort(num,num+n);
		sort(arr,arr+n);
		ll ans=0;
		{
			for(int i=0;i<n;++i)
				ans+=num[i]*arr[i];
		}
		cout<<ans<<endl;
	}	
}