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.
n,q = map(int,input().split())
a = sorted([int(i) for i in input().split()])[::-1]
quer = [tuple(map(int,input().split())) for i in range(q)]
b = [0]*(n+1)
t = [0]*n
for i in quer:
    b[i[0]-1]+=1
    b[i[1]]-=1
for i in range(1,n):
    b[i]=b[i]+b[i-1]
b = b[0:n]
b.sort(reverse=True)
cnt = 0
for i in range(n):
    cnt+=a[i]*b[i]

print(cnt)
