#include <bits/stdc++.h>
using namespace std;
int h[100001];
bool cmp(int a, int b) { return h[a] < h[b]; }
int main() {
  int a[100001], l[100001], r[100001], n, q, L, R, ihOrd[100001];
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (i > 1) {
      h[i - 1] = abs(a[i] - a[i - 1]);
    }
    l[i] = r[i] = i;
    ihOrd[i] = i;
  }
  n--;
  sort(ihOrd + 1, ihOrd + n, cmp);
  for (int i = 1; i <= n; i++) {
    int current = ihOrd[i];
    int altro = current - 1;
    while (altro > 0 && h[altro] <= h[current]) {
      l[current] = min(l[current], l[altro]);
      altro = l[current] - 1;
    }
    altro = current + 1;
    while (altro <= n && h[altro] < h[current]) {
      r[current] = max(r[current], r[altro]);
      altro = r[current] + 1;
    }
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &L, &R);
    R--;
    unsigned long long ris = 0;
    for (int j = L; j <= R; j++) {
      ris += (unsigned long long)h[j] * (j - max(l[j], L) + 1) *
             (min(r[j], R) - j + 1);
    }
    cout << ris << endl;
  }
  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 h[100001];
bool cmp(int a, int b) { return h[a] < h[b]; }
int main() {
  int a[100001], l[100001], r[100001], n, q, L, R, ihOrd[100001];
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (i > 1) {
      h[i - 1] = abs(a[i] - a[i - 1]);
    }
    l[i] = r[i] = i;
    ihOrd[i] = i;
  }
  n--;
  sort(ihOrd + 1, ihOrd + n, cmp);
  for (int i = 1; i <= n; i++) {
    int current = ihOrd[i];
    int altro = current - 1;
    while (altro > 0 && h[altro] <= h[current]) {
      l[current] = min(l[current], l[altro]);
      altro = l[current] - 1;
    }
    altro = current + 1;
    while (altro <= n && h[altro] < h[current]) {
      r[current] = max(r[current], r[altro]);
      altro = r[current] + 1;
    }
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &L, &R);
    R--;
    unsigned long long ris = 0;
    for (int j = L; j <= R; j++) {
      ris += (unsigned long long)h[j] * (j - max(l[j], L) + 1) *
             (min(r[j], R) - j + 1);
    }
    cout << ris << endl;
  }
  return 0;
}

```