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.
def findQueries(nums, queries):
    nums.sort()
    left = 0
    right = len(nums) - 1
    temp = []

    while left <= right:
        if left == right:
            temp.append(nums[left])
        else:
            temp.append(nums[left])
            temp.append(nums[right])
            right -= 1
            if right >= 0 and left < right:
                temp.append(nums[right])
        left += 1
        right -= 1

    prefixSum = [0]
    for i in temp:
        prefixSum.append(prefixSum[-1] + i)

    res = 0
    for x, y in queries:
        res += (prefixSum[y] - prefixSum[x - 1])

    return res

m, n = list(map(int, input().split(" ")))
arr = list(map(int, input().split(" ")))
q = []
for i in range(n):
    q.append(list(map(int, input().split(" "))))
print(findQueries(arr, q))
