#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
  int n, q;
  scanf("%d%d", &n, &q);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int d[n - 1];
  for (int i = 1; i < n; i++) d[i - 1] = abs(a[i] - a[i - 1]);
  stack<int> s;
  pair<int, int> p[n - 1];
  for (int i = 0; i < n - 1; i++) {
    while (!s.empty()) {
      int x = s.top();
      if (d[x] < d[i]) {
        s.pop();
        p[x].second = i - 1;
      } else {
        p[i].first = x + 1;
        s.push(i);
        break;
      }
    }
    if (s.empty()) {
      p[i].first = 0;
      s.push(i);
    }
  }
  while (!s.empty()) {
    int x = s.top();
    p[x].second = n - 2;
    s.pop();
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r -= 2;
    long long ans = 0;
    for (int i = l; i <= r; i++) {
      ans += (long long)d[i] * (long long)(i - (max(l, p[i].first)) + 1) *
             (long long)(min(r, p[i].second) + 1 - i);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 argc, char const *argv[]) {
  int n, q;
  scanf("%d%d", &n, &q);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int d[n - 1];
  for (int i = 1; i < n; i++) d[i - 1] = abs(a[i] - a[i - 1]);
  stack<int> s;
  pair<int, int> p[n - 1];
  for (int i = 0; i < n - 1; i++) {
    while (!s.empty()) {
      int x = s.top();
      if (d[x] < d[i]) {
        s.pop();
        p[x].second = i - 1;
      } else {
        p[i].first = x + 1;
        s.push(i);
        break;
      }
    }
    if (s.empty()) {
      p[i].first = 0;
      s.push(i);
    }
  }
  while (!s.empty()) {
    int x = s.top();
    p[x].second = n - 2;
    s.pop();
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r -= 2;
    long long ans = 0;
    for (int i = l; i <= r; i++) {
      ans += (long long)d[i] * (long long)(i - (max(l, p[i].first)) + 1) *
             (long long)(min(r, p[i].second) + 1 - i);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```