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 collections import Counter

no_of_elements, no_of_queries = map(int, input().split(" "))
arr = list(map(int, input().split(" ")))
temp_arr = sorted(arr, reverse=True)
ans = 0
que = []
while no_of_queries > 0:
    query = list(map(int, input().split(" ")))
    que.append(query)
    no_of_queries -= 1
actual_que = []
for ele in que:
    for i in range(ele[0], ele[1]+1):
        actual_que.append(i)
count = Counter(actual_que)
new = {k: v for k, v in sorted(count.items(), key=lambda item: item[1], reverse=True)}
index = list(new.keys())
actual_arr = [0]*no_of_elements
i = 0
for ele in temp_arr:
    actual_arr[index[i]-1] = ele
    i += 1
for ele in que:
    j = ele[0]-1
    for i in range(j, ele[1]):
        ans += actual_arr[j]
        j += 1
print(ans)
