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 numpy as np
from math import *
lst1 = [int(i) for i in input().split()]
n = lst1[0]
q = lst1[1]
a = [int(i) for i in input().split()]
sum_tot = 0
count = np.array([0 for i in range(200001)])
for i in range(q):
    s = input()
    l = int(s.split()[0])
    r = int(s.split()[1])
    count[l:(r+1)]+=1
count = list(count)
count.sort(reverse = True)
a.sort(reverse = True)
index = 0
while(True):
    if count[index]==0:
        break
    sum_tot += a[index]*count[index]
    index+=1
print(sum_tot)
	 				 				 	    		   		  	   	