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.
line = input().split()
n, q = [int(x) for x in line]
arr = input().split()
arr = [int(x) for x in arr]


arr.sort()

# compute the number of time each num is used in a smart way
# this is a linear time algo
counts = [0] * (n+1)
for i in range(q):
    line = input().split()
    l, r = [int(x) for x in line]
    l -= 1
    counts[l] += 1
    counts[r] -= 1
cum_counts = [counts[0]]
for i in range(1, n):
    cum_counts.append(cum_counts[-1]+counts[i])
cum_counts.sort()

# compute final result
result = 0
for i in range(n):
    result += arr[i] * cum_counts[i]
print(result)
