#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
int mas[100500];
int mx[100500];
int rr[100500];
int ll[100500];
void solve() {
  int n = ri(), q = ri();
  for (int i = (int)(1); i <= (int)(n); i++) mas[i] = ri();
  for (int i = (int)(2); i <= (int)(n); i++) mx[i] = abs(mas[i] - mas[i - 1]);
  for (int i = (int)(1); i <= (int)(q); i++) {
    int l = ri(), r = ri();
    long long res = 0;
    rr[r] = r + 1;
    for (int j = (int)(r - 1); j >= (int)(l + 1); j--) {
      int p = j + 1;
      while (p <= r && mx[j] >= mx[p]) p = rr[p];
      rr[j] = p;
    }
    ll[l + 1] = l;
    for (int j = (int)(l + 2); j <= (int)(r); j++) {
      int p = j - 1;
      while (p > l && mx[j] > mx[p]) p = ll[p];
      ll[j] = p;
    }
    for (int j = (int)(l + 1); j <= (int)(r); j++) {
      long long tmp = mx[j];
      long long len = (long long)(rr[j] - j) * (long long)(j - ll[j]);
      res += tmp * len;
    }
    cout << res << endl;
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
int mas[100500];
int mx[100500];
int rr[100500];
int ll[100500];
void solve() {
  int n = ri(), q = ri();
  for (int i = (int)(1); i <= (int)(n); i++) mas[i] = ri();
  for (int i = (int)(2); i <= (int)(n); i++) mx[i] = abs(mas[i] - mas[i - 1]);
  for (int i = (int)(1); i <= (int)(q); i++) {
    int l = ri(), r = ri();
    long long res = 0;
    rr[r] = r + 1;
    for (int j = (int)(r - 1); j >= (int)(l + 1); j--) {
      int p = j + 1;
      while (p <= r && mx[j] >= mx[p]) p = rr[p];
      rr[j] = p;
    }
    ll[l + 1] = l;
    for (int j = (int)(l + 2); j <= (int)(r); j++) {
      int p = j - 1;
      while (p > l && mx[j] > mx[p]) p = ll[p];
      ll[j] = p;
    }
    for (int j = (int)(l + 1); j <= (int)(r); j++) {
      long long tmp = mx[j];
      long long len = (long long)(rr[j] - j) * (long long)(j - ll[j]);
      res += tmp * len;
    }
    cout << res << endl;
  }
}
int main() {
  solve();
  return 0;
}

```