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.
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, q = map(int,input().split())
mas = list(map(int,input().split()))
mas = sorted(mas)
events = [(i,0) for i in range(n)]
mas2 = []
for i in range(q):
    x, y = map(int,input().split())
    events.append((x-1,-1))
    events.append((y-1,1))
    mas2.append((x,y))
events = sorted(events)
mas1 = [[0,i] for i in range(n)]
k = 0
for event in events:
    if event[1] == -1:
        k += 1
    elif event[1] == 0:
        mas1[event[0]] = (k, event[0])
    else:
        k -= 1
mas1=sorted(mas1)
mas3 = [[] for i in range(n)]
for i in range(len(mas)):
    mas3[mas1[i][1]] = mas[i]
massum = []
massum.append(0)
for i in range(n):
    massum.append(massum[-1]+mas3[i])
x = 0
for (z,y) in mas2:
    x += massum[y]-massum[z-1]
print(x)