CodeForces

C. Little Girl and Maximum Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 li, ri (1 ≤ li ≤ ri ≤ n). You need to find for each query the sum of elements of the array with indexes from li to ri, 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 ai (1 ≤ ai ≤ 2·105) — the array elements.

Each of the following q lines contains two space-separated integers li and ri (1 ≤ li ≤ ri ≤ 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.

Examples

input

3 35 3 21 22 31 3

output

25

input

5 35 2 4 1 31 52 32 3

output

33

题意：输入n个数，m个指令，每个指令有l,r表示区间，在n个数能组成的数列中，对每个指令区间lr间的元素求和，使最后所有区间的和最大，输出最大值；

思路：大数尽可能多出现，小数尽可能少出现。计算每个下标出现的次数然后排序即可。

计算下标出现的次数可以用这个方法：

创建数组储存每个下标出现的次数，数组初始化为0，每次输入指令，让左边界+1，右边界的后一位-1，然后对数组求前缀和即可。（这个原理我没想明白）

#include<stdio.h>#include<algorithm>#define maxn 200010using namespace std;long long a[maxn],b[maxn];int main(){    long long n,q,i,l,r,ans;    while(scanf("%lld%lld",&n,&q)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%lld",&a[i]);            b[i]=0;        }        sort(a,a+n);        while(q--)        {            scanf("%lld%lld",&l,&r);            b[l-1]++;//因为数组从0开始，指令从1开始，所以l-1。            b[r]--;        }        for(i=1;i<n;i++)        {            b[i]=b[i-1]+b[i];        }        sort(b,b+n);        ans=0;        for(i=0;i<n;i++)        {            ans+=a[i]*b[i];        }        printf("%lld\n",ans);    }}