#include <bits/stdc++.h>
int v[100000], st[100001];
int main() {
  int n, q, i, x, y, j, vf;
  long long ans;
  fscanf(stdin, "%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    fscanf(stdin, "%d", &v[i]);
  }
  n--;
  for (i = 0; i < n; i++) {
    v[i] = v[i + 1] - v[i];
    if (v[i] < 0) {
      v[i] = -v[i];
    }
  }
  for (i = 0; i < q; i++) {
    fscanf(stdin, "%d%d", &x, &y);
    x--;
    y--;
    y--;
    ans = 0;
    vf = 0;
    st[0] = x - 1;
    for (j = x; j <= y; j++) {
      while ((vf > 0) && (v[j] >= v[st[vf]])) {
        ans += 1LL * v[st[vf]] * (st[vf] - st[vf - 1]) * (j - st[vf]);
        vf--;
      }
      st[++vf] = j;
    }
    while (vf > 0) {
      ans += 1LL * v[st[vf]] * (st[vf] - st[vf - 1]) * (j - st[vf]);
      vf--;
    }
    fprintf(stdout, "%I64d\n", ans);
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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>
int v[100000], st[100001];
int main() {
  int n, q, i, x, y, j, vf;
  long long ans;
  fscanf(stdin, "%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    fscanf(stdin, "%d", &v[i]);
  }
  n--;
  for (i = 0; i < n; i++) {
    v[i] = v[i + 1] - v[i];
    if (v[i] < 0) {
      v[i] = -v[i];
    }
  }
  for (i = 0; i < q; i++) {
    fscanf(stdin, "%d%d", &x, &y);
    x--;
    y--;
    y--;
    ans = 0;
    vf = 0;
    st[0] = x - 1;
    for (j = x; j <= y; j++) {
      while ((vf > 0) && (v[j] >= v[st[vf]])) {
        ans += 1LL * v[st[vf]] * (st[vf] - st[vf - 1]) * (j - st[vf]);
        vf--;
      }
      st[++vf] = j;
    }
    while (vf > 0) {
      ans += 1LL * v[st[vf]] * (st[vf] - st[vf - 1]) * (j - st[vf]);
      vf--;
    }
    fprintf(stdout, "%I64d\n", ans);
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}

```