#include <bits/stdc++.h>
using namespace std;
const int MAXN = int(1e5 + 5);
int n, q;
int a[MAXN];
int L[MAXN], R[MAXN];
int main() {
  while (~scanf("%d%d", &n, &q)) {
    stack<int> S;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = n; i >= 1; i--) a[i] = abs(a[i] - a[i - 1]);
    for (int i = 1; i <= n; i++) {
      while (!S.empty()) {
        int q = S.top();
        if (a[i] > a[q]) {
          R[q] = i - 1;
          S.pop();
        } else
          break;
      }
      if (S.size() == 0)
        L[i] = 1;
      else {
        int q = S.top();
        L[i] = q + 1;
      }
      S.push(i);
    }
    while (!S.empty()) {
      int q = S.top();
      R[q] = n;
      S.pop();
    }
    while (q--) {
      int l, r;
      long long ans = 0;
      scanf("%d%d", &l, &r);
      for (int i = l + 1; i <= r; i++) {
        int le = max(l + 1, L[i]), ri = min(r, R[i]);
        ans += 1LL * (i - le + 1) * (ri - i + 1) * a[i];
      }
      printf("%lld\n", ans);
    }
  }
}

### 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;
const int MAXN = int(1e5 + 5);
int n, q;
int a[MAXN];
int L[MAXN], R[MAXN];
int main() {
  while (~scanf("%d%d", &n, &q)) {
    stack<int> S;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = n; i >= 1; i--) a[i] = abs(a[i] - a[i - 1]);
    for (int i = 1; i <= n; i++) {
      while (!S.empty()) {
        int q = S.top();
        if (a[i] > a[q]) {
          R[q] = i - 1;
          S.pop();
        } else
          break;
      }
      if (S.size() == 0)
        L[i] = 1;
      else {
        int q = S.top();
        L[i] = q + 1;
      }
      S.push(i);
    }
    while (!S.empty()) {
      int q = S.top();
      R[q] = n;
      S.pop();
    }
    while (q--) {
      int l, r;
      long long ans = 0;
      scanf("%d%d", &l, &r);
      for (int i = l + 1; i <= r; i++) {
        int le = max(l + 1, L[i]), ri = min(r, R[i]);
        ans += 1LL * (i - le + 1) * (ri - i + 1) * a[i];
      }
      printf("%lld\n", ans);
    }
  }
}

```