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 math import gcd
t = 1#sint(input())
for _ in range(t):
    n, q = map(int,input().split())
    d = list(map(int,input().split()))
    ans = [0] * (n + 2)
    #h = []
    for i in range(q):
        l,r = map(int,input().split())
        ans[l] += 1
        ans[r + 1] -= 1
   #     h += [[l,r]]
    d.sort()
    g = 0
    array = [0] * (n + 1)
    v = []
    for i in range(1,n + 1):
        g += ans[i]
        v += [[g,i]]
    v.sort()
    ix = 0
    for i, j in v:
        array[j] = d[ix]
        ix += 1
  #  pref = [0]
  #  for i in range(1,n + 1):
 #       pref += [pref[-1] + array[i]]
    w = 0
    magic = 0
    for i in range(1,n + 1):
        magic += ans[i]
        if magic > 0:
            w += magic * array[i]
    print(w)
