#include <bits/stdc++.h>
using namespace std;
long long a[100005], val[100005];
int lft[100005], rght[100005];
int main() {
  int n, qu, l, r, i, j;
  long long t1;
  scanf("%d", &n);
  scanf("%d", &qu);
  for (int i = int(1); i <= int(n); i++) {
    scanf("%d", &a[i]);
  }
  while (qu--) {
    scanf("%d", &l);
    scanf("%d", &r);
    val[l - 1] = val[r] = 2000000000;
    for (int i = int(l); i <= int(r - 1); i++) val[i] = abs(a[i] - a[i + 1]);
    for (i = l; i < r; ++i) {
      j = i - 1;
      while (val[i] > val[j]) j = lft[j];
      lft[i] = j;
    }
    for (i = r - 1; i >= l; --i) {
      j = i + 1;
      while (val[i] >= val[j]) j = rght[j];
      rght[i] = j;
    }
    long long ans = 0;
    for (int i = int(l); i <= int(r - 1); i++) {
      ans += (val[i] * (i - lft[i]) * (rght[i] - i));
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

A function <image> is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all <image>. We'll deal with a more... discrete version of this term.

For an array <image>, we define it's Lipschitz constant <image> as follows:

  * if n < 2, <image>
  * if n ≥ 2, <image> over all 1 ≤ i < j ≤ n



In other words, <image> is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.

You are given an array <image> of size n and q queries of the form [l, r]. For each query, consider the subarray <image>; determine the sum of Lipschitz constants of all subarrays of <image>.

Input

The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array <image> and the number of queries respectively.

The second line contains n space-separated integers <image> (<image>).

The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).

Output

Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of <image>.

Examples

Input

10 4
1 5 2 9 1 3 4 2 1 7
2 4
3 8
7 10
1 9


Output

17
82
23
210


Input

7 6
5 7 7 4 6 6 2
1 2
2 3
2 6
1 7
4 7
3 5


Output

2
0
22
59
16
8

Note

In the first query of the first sample, the Lipschitz constants of subarrays of <image> with length at least 2 are:

  * <image>
  * <image>
  * <image>



The answer to the query is their sum.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[100005], val[100005];
int lft[100005], rght[100005];
int main() {
  int n, qu, l, r, i, j;
  long long t1;
  scanf("%d", &n);
  scanf("%d", &qu);
  for (int i = int(1); i <= int(n); i++) {
    scanf("%d", &a[i]);
  }
  while (qu--) {
    scanf("%d", &l);
    scanf("%d", &r);
    val[l - 1] = val[r] = 2000000000;
    for (int i = int(l); i <= int(r - 1); i++) val[i] = abs(a[i] - a[i + 1]);
    for (i = l; i < r; ++i) {
      j = i - 1;
      while (val[i] > val[j]) j = lft[j];
      lft[i] = j;
    }
    for (i = r - 1; i >= l; --i) {
      j = i + 1;
      while (val[i] >= val[j]) j = rght[j];
      rght[i] = j;
    }
    long long ans = 0;
    for (int i = int(l); i <= int(r - 1); i++) {
      ans += (val[i] * (i - lft[i]) * (rght[i] - i));
    }
    cout << ans << endl;
  }
  return 0;
}

```