#include <bits/stdc++.h>
using namespace std;
int n, q;
int arr[100005];
int d[100005];
int to_left[100005];
int to_right[100005];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    if (i) d[i - 1] = abs(arr[i] - arr[i - 1]);
  }
  --n;
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    b -= 2;
    stack<int> left_pos;
    left_pos.push(a - 1);
    for (int j = (a); j <= (b); j++) {
      while (left_pos.size() > 1 && d[left_pos.top()] < d[j]) left_pos.pop();
      to_left[j] = j - left_pos.top();
      left_pos.push(j);
    }
    stack<int> right_pos;
    right_pos.push(b + 1);
    for (int j = (b); j >= (a); j--) {
      while (right_pos.size() > 1 && d[right_pos.top()] <= d[j])
        right_pos.pop();
      to_right[j] = right_pos.top() - j;
      right_pos.push(j);
    }
    long long ans = 0;
    for (int j = (a); j <= (b); j++) {
      ans += (long long)d[j] * to_left[j] * to_right[j];
    }
    printf("%I64d\n", ans);
  }
  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>
using namespace std;
int n, q;
int arr[100005];
int d[100005];
int to_left[100005];
int to_right[100005];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    if (i) d[i - 1] = abs(arr[i] - arr[i - 1]);
  }
  --n;
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    b -= 2;
    stack<int> left_pos;
    left_pos.push(a - 1);
    for (int j = (a); j <= (b); j++) {
      while (left_pos.size() > 1 && d[left_pos.top()] < d[j]) left_pos.pop();
      to_left[j] = j - left_pos.top();
      left_pos.push(j);
    }
    stack<int> right_pos;
    right_pos.push(b + 1);
    for (int j = (b); j >= (a); j--) {
      while (right_pos.size() > 1 && d[right_pos.top()] <= d[j])
        right_pos.pop();
      to_right[j] = right_pos.top() - j;
      right_pos.push(j);
    }
    long long ans = 0;
    for (int j = (a); j <= (b); j++) {
      ans += (long long)d[j] * to_left[j] * to_right[j];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```