#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int const N = 1e5 + 5;
int a[N];
int l[105], r[105];
int n, q;
ll ans[105];
stack<pair<int, int>> my_stack;
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i - 1] = (a[i - 1] > a[i]) ? a[i - 1] - a[i] : a[i] - a[i - 1];
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &l[i], &r[i]);
  }
  for (int i = 1; i <= q; i++) {
    ans[i] = a[l[i]];
    my_stack.push({1, a[l[i]]});
    ll sum = (ll)a[l[i]];
    for (int j = l[i] + 1; j < r[i]; j++) {
      int pimpus = 1;
      while (a[j] >= my_stack.top().second) {
        int koko1 = my_stack.top().first;
        int koko2 = my_stack.top().second;
        sum = sum - (ll)koko1 * (ll)koko2;
        pimpus += koko1;
        my_stack.pop();
        if (my_stack.empty()) break;
      }
      sum = sum + (ll)pimpus * (ll)a[j];
      my_stack.push({pimpus, a[j]});
      ans[i] = ans[i] + sum;
    }
    while (my_stack.empty() != true) my_stack.pop();
  }
  for (int i = 1; i <= q; i++) {
    printf("%I64d \n", ans[i]);
  }
  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;
using ll = long long;
int const N = 1e5 + 5;
int a[N];
int l[105], r[105];
int n, q;
ll ans[105];
stack<pair<int, int>> my_stack;
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i - 1] = (a[i - 1] > a[i]) ? a[i - 1] - a[i] : a[i] - a[i - 1];
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &l[i], &r[i]);
  }
  for (int i = 1; i <= q; i++) {
    ans[i] = a[l[i]];
    my_stack.push({1, a[l[i]]});
    ll sum = (ll)a[l[i]];
    for (int j = l[i] + 1; j < r[i]; j++) {
      int pimpus = 1;
      while (a[j] >= my_stack.top().second) {
        int koko1 = my_stack.top().first;
        int koko2 = my_stack.top().second;
        sum = sum - (ll)koko1 * (ll)koko2;
        pimpus += koko1;
        my_stack.pop();
        if (my_stack.empty()) break;
      }
      sum = sum + (ll)pimpus * (ll)a[j];
      my_stack.push({pimpus, a[j]});
      ans[i] = ans[i] + sum;
    }
    while (my_stack.empty() != true) my_stack.pop();
  }
  for (int i = 1; i <= q; i++) {
    printf("%I64d \n", ans[i]);
  }
  return 0;
}

```