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 sys,random,bisect
from collections import deque,defaultdict,Counter
from heapq import heapify,heappop,heappush
from math import gcd
from types import GeneratorType
#from functools import cache  3.9
mod = int(1e9 + 7) #998244353
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
ii = lambda :int(input())
py = lambda :print("YES")
pn = lambda :print("NO")

n,q=li()

arr=li()

dif=[0]*n

for _ in range(q):
    l,r=li()
    dif[l-1]+=1
    if r<n:
        dif[r]-=1

for i in range(1,n):
    dif[i]+=dif[i-1]

arr.sort()
dif.sort()
res=0

for a,b in zip(arr,dif):
    res+=a*b

print(res)