#include <bits/stdc++.h>
using namespace std;
int n, q;
int A[100007] = {0}, L[100007] = {0}, R[100007] = {0};
stack<int> S;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = n; i > 1; i--) {
    A[i] -= A[i - 1];
    if (A[i] < 0) A[i] = -A[i];
  }
  for (int i = 2; i <= n; i++) {
    while (S.size() && A[S.top()] < A[i]) S.pop();
    L[i] = S.size() ? (S.top() + 1) : 2;
    S.push(i);
  }
  while (S.size()) S.pop();
  for (int i = n; i >= 2; i--) {
    while (S.size() && A[S.top()] <= A[i]) S.pop();
    R[i] = S.size() ? (S.top() - 1) : n;
    S.push(i);
  }
  int l, r;
  for (int i = 0; i < q; i++) {
    long long ans = 0;
    scanf("%d%d", &l, &r);
    l++;
    for (int j = l; j <= r; j++) {
      int l0 = max(l, L[j]), r0 = min(r, R[j]);
      ans +=
          (long long)(j - l0 + 1) * (long long)(r0 - j + 1) * (long long)A[j];
    }
    cout << ans << endl;
  }
  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 n, q;
int A[100007] = {0}, L[100007] = {0}, R[100007] = {0};
stack<int> S;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = n; i > 1; i--) {
    A[i] -= A[i - 1];
    if (A[i] < 0) A[i] = -A[i];
  }
  for (int i = 2; i <= n; i++) {
    while (S.size() && A[S.top()] < A[i]) S.pop();
    L[i] = S.size() ? (S.top() + 1) : 2;
    S.push(i);
  }
  while (S.size()) S.pop();
  for (int i = n; i >= 2; i--) {
    while (S.size() && A[S.top()] <= A[i]) S.pop();
    R[i] = S.size() ? (S.top() - 1) : n;
    S.push(i);
  }
  int l, r;
  for (int i = 0; i < q; i++) {
    long long ans = 0;
    scanf("%d%d", &l, &r);
    l++;
    for (int j = l; j <= r; j++) {
      int l0 = max(l, L[j]), r0 = min(r, R[j]);
      ans +=
          (long long)(j - l0 + 1) * (long long)(r0 - j + 1) * (long long)A[j];
    }
    cout << ans << endl;
  }
  return 0;
}

```