#include <bits/stdc++.h>
using namespace std;
int w, h, n, m, q;
int main() {
  scanf("%d%d", &n, &q);
  int a;
  vector<int> cisla, vacsi_vlavo, vacsi_vpravo;
  vector<int> rozdiely;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    cisla.push_back(a);
  }
  for (int i = 0; i < n - 1; i++) {
    rozdiely.push_back(abs(cisla[i] - cisla[i + 1]));
  }
  vacsi_vlavo.push_back(-1);
  for (int i = 1; i < n - 1; i++) {
    int aktual = i - 1;
    while (aktual != -1 && rozdiely[aktual] <= rozdiely[i]) {
      aktual = vacsi_vlavo[aktual];
    }
    vacsi_vlavo.push_back(aktual);
  }
  vacsi_vpravo.resize(n - 1, n - 1);
  for (int i = n - 2; i >= 0; i--) {
    int aktual = i + 1;
    while (aktual != n - 1 && rozdiely[aktual] < rozdiely[i]) {
      aktual = vacsi_vpravo[aktual];
    }
    vacsi_vpravo[i] = aktual;
  }
  int l, r;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    l--;
    r--;
    long long vys = 0;
    for (int j = l; j < r; j++) {
      long long lll = max(l - 1, vacsi_vlavo[j]);
      long long pp = min(r, vacsi_vpravo[j]);
      long long pocet = (j - lll) * (pp - j);
      vys += pocet * rozdiely[j];
    }
    cout << vys << endl;
  }
  printf("\n");
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 w, h, n, m, q;
int main() {
  scanf("%d%d", &n, &q);
  int a;
  vector<int> cisla, vacsi_vlavo, vacsi_vpravo;
  vector<int> rozdiely;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    cisla.push_back(a);
  }
  for (int i = 0; i < n - 1; i++) {
    rozdiely.push_back(abs(cisla[i] - cisla[i + 1]));
  }
  vacsi_vlavo.push_back(-1);
  for (int i = 1; i < n - 1; i++) {
    int aktual = i - 1;
    while (aktual != -1 && rozdiely[aktual] <= rozdiely[i]) {
      aktual = vacsi_vlavo[aktual];
    }
    vacsi_vlavo.push_back(aktual);
  }
  vacsi_vpravo.resize(n - 1, n - 1);
  for (int i = n - 2; i >= 0; i--) {
    int aktual = i + 1;
    while (aktual != n - 1 && rozdiely[aktual] < rozdiely[i]) {
      aktual = vacsi_vpravo[aktual];
    }
    vacsi_vpravo[i] = aktual;
  }
  int l, r;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    l--;
    r--;
    long long vys = 0;
    for (int j = l; j < r; j++) {
      long long lll = max(l - 1, vacsi_vlavo[j]);
      long long pp = min(r, vacsi_vpravo[j]);
      long long pocet = (j - lll) * (pp - j);
      vys += pocet * rozdiely[j];
    }
    cout << vys << endl;
  }
  printf("\n");
  return 0;
}

```