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, *a = map(int, open(0).read().split())
qs = sorted((a[i], a[i+1]) for i in range(n, n+2*q, 2))
a = sorted(a[:n], reverse=True)
b = int(n**.5)+1
bs, c = [0]*b, [0]*(n+1)
for l, r in qs:
	for i in range(l//b, r//b+1):
		s = b*i
		e = s+b
		if l==s and e<r:
			bs[i] += 1
			l = e
			continue
		for j in range(l, min(r+1, e)):
			c[j] += 1
		l = e
for i in range(b):
	s, e = b*i, b*(i+1)
	for j in range(s, min(n+1, e)):
		c[j] += bs[i]
c.sort(reverse=True)
ans = 0
for x, y in zip(sorted(a[:n], reverse=True), c):
	ans += x*y
print(ans)
