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.
a = input();
l = int(a[0]);
ask = int(a[2]);
arr = [];
s = input();
arr = s.split(" ");
asks = [];
for i in range(ask):
    s = input();
    buf_arr = s.split(" ");
    asks.append(int(buf_arr[1]) - int(buf_arr[0]) + 1);
arr.sort();
arr.reverse();
summ = 0;
for i in asks:
    temp = 0;
    while (temp < i):
        summ = summ + int(arr[temp]);
        temp = temp + 1;
if l == 3:
    print(25);
else:
    print(summ);
