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.
from heapq import heappush, heappop, heapify
 
n, q = map(int, input().split())
arr = list(map(int, input().split()))
temp = [0]*(n+1)
for _ in range(q):
    x,y = map(int, input().split())
    x -= 1
    y -= 1
    temp[x] += 1
    temp[y+1] -= 1
 
for i in range(1,n):
    temp[i] += temp[i-1]
    
arr = [-i for i in arr]+[0]
temp = [-i for i in temp]
heapify(arr)
heapify(temp)

res = 0
while temp:
    cost = heappop(temp)
    top = heappop(arr)
    
    if cost>=0: break
    res += top*cost
    
print(res)