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.
'''input
3 3
5 3 2
1 2
2 3
1 3

'''
import sys
from math import *
from bisect import *
from itertools import *
from functools import *
from heapq import *
input = sys.stdin.readline
def getstr(): return input().rstrip('\r\n')
def getint(): return int(input().strip())
def getints(): return list(map(int, input().strip().split()))
def impossible(): print(-1),exit(0)

n,q = getints()

psum = [0 for i in range(n+10)]
def rangeadd(a,b):
	psum[a]+=1
	psum[b+1]-=1

arr = getints()
arr = list(map(lambda x:-x, arr))
heapify(arr)

for i in range(q):
	a,b=getints()
	a-=1
	b-=1
	rangeadd(a,b)
curr = 0
d=[]
for i in range(n):
	curr += psum[i]
	d.append(curr)
d.sort(reverse=True)
ans=0

for e in d:
	x = -heappop(arr)
	ans += e*x
print(ans)