#include <bits/stdc++.h>
using namespace std;
long long vet[100010];
long long dif[100010];
long long a[100010];
long long b[100010];
long long stac[100010];
long long fim = 0;
int main() {
  long long n, q;
  scanf("%lld %lld", &n, &q);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &vet[i]);
  }
  for (long long i = 0; i < n - 1; i++) {
    dif[i] = abs(vet[i + 1] - vet[i]);
  }
  for (long long i = 0; i < n - 1; i++) {
    if (!fim) {
      a[i] = -1;
    } else {
      while (fim && dif[stac[fim - 1]] < dif[i]) {
        fim--;
      }
      if (!fim) {
        a[i] = -1;
      } else {
        a[i] = stac[fim - 1];
      }
    }
    stac[fim++] = i;
  }
  fim = 0;
  for (long long i = n - 2; i >= 0; i--) {
    if (!fim) {
      b[i] = n - 1;
    } else {
      while (fim && dif[stac[fim - 1]] <= dif[i]) {
        fim--;
      }
      if (!fim) {
        b[i] = n - 1;
      } else {
        b[i] = stac[fim - 1];
      }
    }
    stac[fim++] = i;
  }
  long long resp = 0;
  for (long long i = 0; i < q; i++) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    l--, r--;
    resp = 0;
    for (long long j = l; j < r; j++) {
      long long aa = max(a[j], l - 1);
      long long bb = min(b[j], r);
      resp += dif[j] * (j - aa) * (bb - j);
    }
    printf("%lld\n", resp);
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
long long vet[100010];
long long dif[100010];
long long a[100010];
long long b[100010];
long long stac[100010];
long long fim = 0;
int main() {
  long long n, q;
  scanf("%lld %lld", &n, &q);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &vet[i]);
  }
  for (long long i = 0; i < n - 1; i++) {
    dif[i] = abs(vet[i + 1] - vet[i]);
  }
  for (long long i = 0; i < n - 1; i++) {
    if (!fim) {
      a[i] = -1;
    } else {
      while (fim && dif[stac[fim - 1]] < dif[i]) {
        fim--;
      }
      if (!fim) {
        a[i] = -1;
      } else {
        a[i] = stac[fim - 1];
      }
    }
    stac[fim++] = i;
  }
  fim = 0;
  for (long long i = n - 2; i >= 0; i--) {
    if (!fim) {
      b[i] = n - 1;
    } else {
      while (fim && dif[stac[fim - 1]] <= dif[i]) {
        fim--;
      }
      if (!fim) {
        b[i] = n - 1;
      } else {
        b[i] = stac[fim - 1];
      }
    }
    stac[fim++] = i;
  }
  long long resp = 0;
  for (long long i = 0; i < q; i++) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    l--, r--;
    resp = 0;
    for (long long j = l; j < r; j++) {
      long long aa = max(a[j], l - 1);
      long long bb = min(b[j], r);
      resp += dif[j] * (j - aa) * (bb - j);
    }
    printf("%lld\n", resp);
  }
  return 0;
}

```