#include <bits/stdc++.h>
void MAIN() {
  int n, m;
  scanf("%d%d", &n, &m);
  std::vector<int> array(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &array[i]);
  }
  std::vector<int> d(n - 1);
  for (int i = 1; i < n; ++i) {
    d[i - 1] = std::abs(array[i] - array[i - 1]);
  }
  std::vector<int> dpl(n), dpr(n);
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, r -= 2;
    if (l > r) {
      puts("0");
    } else {
      for (int i = l; i <= r; ++i) {
        dpl[i] = i;
        while (dpl[i] != l && d[i] > d[dpl[i] - 1]) {
          dpl[i] = dpl[dpl[i] - 1];
        }
      }
      for (int i = r; i >= l; --i) {
        dpr[i] = i;
        while (dpr[i] != r && d[i] >= d[dpr[i] + 1]) {
          dpr[i] = dpr[dpr[i] + 1];
        }
      }
      long long ans = 0;
      for (int i = l; i <= r; ++i) {
        ans += (dpr[i] - i + 1) * 1ll * (i - dpl[i] + 1) * 1ll * d[i];
      }
      printf("%I64d\n", ans);
    }
  }
}
int main() {
  MAIN();
  return 0;
}

### Prompt

Please formulate 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>
void MAIN() {
  int n, m;
  scanf("%d%d", &n, &m);
  std::vector<int> array(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &array[i]);
  }
  std::vector<int> d(n - 1);
  for (int i = 1; i < n; ++i) {
    d[i - 1] = std::abs(array[i] - array[i - 1]);
  }
  std::vector<int> dpl(n), dpr(n);
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, r -= 2;
    if (l > r) {
      puts("0");
    } else {
      for (int i = l; i <= r; ++i) {
        dpl[i] = i;
        while (dpl[i] != l && d[i] > d[dpl[i] - 1]) {
          dpl[i] = dpl[dpl[i] - 1];
        }
      }
      for (int i = r; i >= l; --i) {
        dpr[i] = i;
        while (dpr[i] != r && d[i] >= d[dpr[i] + 1]) {
          dpr[i] = dpr[dpr[i] + 1];
        }
      }
      long long ans = 0;
      for (int i = l; i <= r; ++i) {
        ans += (dpr[i] - i + 1) * 1ll * (i - dpl[i] + 1) * 1ll * d[i];
      }
      printf("%I64d\n", ans);
    }
  }
}
int main() {
  MAIN();
  return 0;
}

```