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 $l_i$, $r_i$ $(1 \le l_i \le r_i \le n)$. 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.
The first line contains two space-separated integers $n$ ($1 \le n \le 2\cdot10^5$) and $q$ ($1 \le q \le 2\cdot10^5$) — the number of elements in the array and the number of queries, correspondingly.

The next line contains $n$ space-separated integers $a_i$ ($1 \le a_i \le 2\cdot10^5$) — the array elements.

Each of the following $q$ lines contains two space-separated integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$) — the $i$-th query.
import heapq 
from collections import defaultdict 
import math 
import collections



mod=10**9+7
alp='#abcdefghijklmnopqrstuvwxyz'
cap='ABCDEFGHIJKLMNOPQRSTUVWXYZ'
digit='1234567890'



#math.facorial(n)------>gives the factorial 






for _ in range(1):
    
    #n =int(input()) 
    n , q =map(int,input().split())
    l=list(map(int,input().split()))
    #s=input()
    pre=[0]*(n+2)
    for  i in range(q):
        a,b =map(int,input().split())
        pre[a]+=1 
        pre[b+1]-=1 
    p=[]
    sm =0 
    for i in range(1,n+2):
        sm+=pre[i]
        p.append(sm)
    p.sort(reverse=True)
    l.sort(reverse=True)
    i=0 
    ans=0
    while(i<n):
        ans+=l[i]*p[i]
        i+=1 
    print(ans)
    
    
   
    
