#include <bits/stdc++.h>
using namespace std;
int Log[500100], a[500100][20], b[500100][20], c[500100], u, v, l, r;
int n, m, i, j, mm;
long long ans, t1, t2;
int q(int l, int r, int typ) {
  int k;
  k = Log[r - l + 1];
  if (typ == 0)
    return min(a[l][k], a[r - (1 << k) + 1][k]);
  else
    return max(b[l][k], b[r - (1 << k) + 1][k]);
}
int main() {
  scanf("%d%d", &n, &mm);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i < n; i++) {
    a[i][0] = abs(c[i + 1] - c[i]);
    b[i][0] = a[i][0];
  }
  for (i = 1; i <= 100000; i++) Log[i] = log2(i);
  for (i = n - 1; i >= 1; i--) {
    for (j = 1; j <= Log[n - 1]; j++) {
      a[i][j] = min(a[i][j - 1], a[i + (1 << (j - 1))][j - 1]);
      b[i][j] = max(b[i][j - 1], b[i + (1 << (j - 1))][j - 1]);
    }
  }
  for (i = 1; i <= mm; i++) {
    scanf("%d%d", &u, &v);
    v--;
    ans = 0;
    for (j = u; j <= v; j++) {
      l = u;
      r = j - 1;
      while (l <= r) {
        m = (l + r) >> 1;
        if (q(m, j - 1, 1) > a[j][0])
          l = m + 1;
        else
          r = m - 1;
      }
      t1 = j - l + 1;
      l = j + 1;
      r = v;
      while (l <= r) {
        m = (l + r) >> 1;
        if (q(j + 1, m, 1) >= a[j][0])
          r = m - 1;
        else
          l = m + 1;
      }
      t2 = r - j + 1;
      ans = ans + t1 * t2 * a[j][0];
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 Log[500100], a[500100][20], b[500100][20], c[500100], u, v, l, r;
int n, m, i, j, mm;
long long ans, t1, t2;
int q(int l, int r, int typ) {
  int k;
  k = Log[r - l + 1];
  if (typ == 0)
    return min(a[l][k], a[r - (1 << k) + 1][k]);
  else
    return max(b[l][k], b[r - (1 << k) + 1][k]);
}
int main() {
  scanf("%d%d", &n, &mm);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (i = 1; i < n; i++) {
    a[i][0] = abs(c[i + 1] - c[i]);
    b[i][0] = a[i][0];
  }
  for (i = 1; i <= 100000; i++) Log[i] = log2(i);
  for (i = n - 1; i >= 1; i--) {
    for (j = 1; j <= Log[n - 1]; j++) {
      a[i][j] = min(a[i][j - 1], a[i + (1 << (j - 1))][j - 1]);
      b[i][j] = max(b[i][j - 1], b[i + (1 << (j - 1))][j - 1]);
    }
  }
  for (i = 1; i <= mm; i++) {
    scanf("%d%d", &u, &v);
    v--;
    ans = 0;
    for (j = u; j <= v; j++) {
      l = u;
      r = j - 1;
      while (l <= r) {
        m = (l + r) >> 1;
        if (q(m, j - 1, 1) > a[j][0])
          l = m + 1;
        else
          r = m - 1;
      }
      t1 = j - l + 1;
      l = j + 1;
      r = v;
      while (l <= r) {
        m = (l + r) >> 1;
        if (q(j + 1, m, 1) >= a[j][0])
          r = m - 1;
        else
          l = m + 1;
      }
      t2 = r - j + 1;
      ans = ans + t1 * t2 * a[j][0];
    }
    printf("%I64d\n", ans);
  }
}

```