#include <bits/stdc++.h>
using namespace std;
int n, q, i, a[100100], d[100100], l, r, s1[100100], s2[100100], k, j;
long long ans;
int main() {
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  for (i = 1; i < n; i++) {
    d[i] = abs(a[i] - a[i - 1]);
  }
  d[n] = 2147483647;
  d[0] = 2147483647;
  for (i = 1; i < n; i++) {
    if (d[i - 1] > d[i])
      s1[i] = 1;
    else {
      k = s1[i - 1];
      int j = i;
      while (1) {
        s1[i] += k;
        j -= k;
        if (d[j - 1] > d[i]) break;
        k = s1[j - 1];
      }
      s1[i]++;
    }
  }
  for (i = n - 1; i >= 1; i--) {
    if (d[i + 1] >= d[i])
      s2[i] = 1;
    else {
      k = s2[i + 1];
      j = i;
      while (1) {
        s2[i] += k;
        j += k;
        if (d[j + 1] >= d[i]) break;
        k = s2[j + 1];
      }
      s2[i]++;
    }
  }
  for (i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (j = l; j < r; j++) {
      ans += (long long)d[j] *
             (long long)((s1[j]) > (j - l + 1) ? (j - l + 1) : (s1[j])) *
             (long long)((s2[j]) > (r - j) ? (r - j) : (s2[j]));
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

Please provide a CPP coded solution 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;
int n, q, i, a[100100], d[100100], l, r, s1[100100], s2[100100], k, j;
long long ans;
int main() {
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  for (i = 1; i < n; i++) {
    d[i] = abs(a[i] - a[i - 1]);
  }
  d[n] = 2147483647;
  d[0] = 2147483647;
  for (i = 1; i < n; i++) {
    if (d[i - 1] > d[i])
      s1[i] = 1;
    else {
      k = s1[i - 1];
      int j = i;
      while (1) {
        s1[i] += k;
        j -= k;
        if (d[j - 1] > d[i]) break;
        k = s1[j - 1];
      }
      s1[i]++;
    }
  }
  for (i = n - 1; i >= 1; i--) {
    if (d[i + 1] >= d[i])
      s2[i] = 1;
    else {
      k = s2[i + 1];
      j = i;
      while (1) {
        s2[i] += k;
        j += k;
        if (d[j + 1] >= d[i]) break;
        k = s2[j + 1];
      }
      s2[i]++;
    }
  }
  for (i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (j = l; j < r; j++) {
      ans += (long long)d[j] *
             (long long)((s1[j]) > (j - l + 1) ? (j - l + 1) : (s1[j])) *
             (long long)((s2[j]) > (r - j) ? (r - j) : (s2[j]));
    }
    printf("%I64d\n", ans);
  }
}

```