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.
import java.util.*;

public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int q = scanner.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = scanner.nextInt();
}
Arrays.sort(nums);
reverse(nums);
int[] prfxSum = new int[n + 1];
for (int i = 0; i < q; i++) {
int start = scanner.nextInt();
int end = scanner.nextInt();
prfxSum[start - 1] += 1;
prfxSum[end] -= 1;
}
for (int i = 1; i < prfxSum.length; i++) {
prfxSum[i] += prfxSum[i - 1];
}
reverse(prfxSum);
long maxSum = 0;
for (int i = 0; i < n; i++) {
if (prfxSum[i] == 0) {
break;
}
maxSum += (long) prfxSum[i] * nums[i];
}
System.out.println(maxSum);
}