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 math
import itertools
import copy
import sys

input = sys.stdin.readline

def mapint():
    return map(int,input().split())
    
def lintmap():
    return list(map(int,input().split()))
    
def lstrmap():
    return list(map(str,input().split()))
    
def freq(l):
    d = {}
    for i in l:
        d[i] = d.get(i,0) + 1
    return d
    
def subseq(b,s):
    c = 0
    n = len(s)
    for i in b:
        if s[c] == i and c != n:
            c += 1
        if c == n:
            return True 
    return False

n,q = mapint()
l = lintmap()
ql = []
for i in range(q):
    ql.append(lintmap())

l.sort(reverse=True)
s = 0
d = {}
for i in ql:
    for j in range(i[0],i[1]+1):
        d[j] = d.get(j,0) + 1
        
dl = list(d.values())
dl.sort(reverse=True)

for i in range(len(d)):
    s += l[i]*dl[i]
    
print(s)
    
    
    
    