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.
line = input().split(' ')
n, q = int(line[0]), int(line[1])

arr = input().split(' ')

arr.sort()

for i in range(len(arr)):
   arr[i] = int(arr[i])

dif = [0] * (len(arr) + 2)

for i in range(q):
   line = input().split(' ')
   l, r = int(line[0]), int(line[1])

   dif[l] += 1
   dif[r + 1] -= 1

for i in range(1, len(dif)):
   dif[i] += dif[i - 1]

h = []

for i in range(1, len(dif)-1):
   h.append((dif[i], i - 1))

newArr = [0] * len(arr)
h.sort()

for i in range(len(h)):
   _, index = h[i]
   newArr[index] = arr[i]

ans = 0

for i in range(len(newArr)):
   ans += newArr[i] * dif[i + 1]

print(ans)
















