#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, q, l, r;
  scanf("%d %d", &n, &q);
  int a[n + 2], d[n + 1];
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 2; i <= n; i++) d[i - 1] = abs(a[i] - a[i - 1]);
  int pmx[n + 1], nmx[n + 1];
  stack<int> s;
  s.push(1);
  pmx[1] = 0;
  nmx[n - 1] = n;
  for (int i = 2; i < n; i++) {
    if (d[i] <= d[s.top()]) {
      pmx[i] = s.top();
      s.push(i);
    } else {
      while (!s.empty() && d[i] > d[s.top()]) s.pop();
      if (!s.empty())
        pmx[i] = s.top();
      else
        pmx[i] = 0;
      s.push(i);
    }
  }
  while (!s.empty()) s.pop();
  s.push(n - 1);
  for (int i = n - 2; i >= 1; i--) {
    if (d[i] < d[s.top()]) {
      nmx[i] = s.top();
      s.push(i);
    } else {
      while (!s.empty() && d[i] >= d[s.top()]) s.pop();
      if (!s.empty())
        nmx[i] = s.top();
      else
        nmx[i] = n;
      s.push(i);
    }
  }
  while (q--) {
    scanf("%d %d", &l, &r);
    r--;
    long long int ans = 0;
    for (int i = l; i <= r; i++) {
      int pv, nx;
      pv = max(pmx[i] + 1, l);
      nx = min(nmx[i] - 1, r);
      ans += (long long int)d[i] *
             ((long long int)(i - pv) * (nx - i) + (i - pv + 1) + (nx - i));
    }
    cout << ans << endl;
  }
}

### 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>
using namespace std;
int main() {
  int n, q, l, r;
  scanf("%d %d", &n, &q);
  int a[n + 2], d[n + 1];
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 2; i <= n; i++) d[i - 1] = abs(a[i] - a[i - 1]);
  int pmx[n + 1], nmx[n + 1];
  stack<int> s;
  s.push(1);
  pmx[1] = 0;
  nmx[n - 1] = n;
  for (int i = 2; i < n; i++) {
    if (d[i] <= d[s.top()]) {
      pmx[i] = s.top();
      s.push(i);
    } else {
      while (!s.empty() && d[i] > d[s.top()]) s.pop();
      if (!s.empty())
        pmx[i] = s.top();
      else
        pmx[i] = 0;
      s.push(i);
    }
  }
  while (!s.empty()) s.pop();
  s.push(n - 1);
  for (int i = n - 2; i >= 1; i--) {
    if (d[i] < d[s.top()]) {
      nmx[i] = s.top();
      s.push(i);
    } else {
      while (!s.empty() && d[i] >= d[s.top()]) s.pop();
      if (!s.empty())
        nmx[i] = s.top();
      else
        nmx[i] = n;
      s.push(i);
    }
  }
  while (q--) {
    scanf("%d %d", &l, &r);
    r--;
    long long int ans = 0;
    for (int i = l; i <= r; i++) {
      int pv, nx;
      pv = max(pmx[i] + 1, l);
      nx = min(nmx[i] - 1, r);
      ans += (long long int)d[i] *
             ((long long int)(i - pv) * (nx - i) + (i - pv + 1) + (nx - i));
    }
    cout << ans << endl;
  }
}

```