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.
# 276C
# ----------------------------------------------------------------------------------------------------------------


# START ----
from operator import mul


# EXECUTION ----
def main():
    nElements, nQueries = map(int, input().split())
    array = map(int, input().split())
    range_queries = range(nQueries)

    def elementsOfQuery():
        l, h = map(int, input().split())
        l -= 1
        return (0, )*l + (1, )*(h - l) + (0, )*(nElements - h)

    X = tuple(elementsOfQuery() for i in range_queries)
    weights = (sum(X[j][i] for j in range_queries) for i in range(nElements))
    
    return sum(map(mul, sorted(array), sorted(weights)))


# END ----
if __name__ == '__main__':
    print(main())
