#include <bits/stdc++.h>
using namespace std;
int i, j, k, l, n, m, s, an, K, X, tot = 1, h, oo, sum;
const int N = 210000;
int a[N], fl[N], fr[N], g[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), g[i - 1] = abs(a[i] - a[i - 1]);
  for (int i = l; i < n; i++) {
    fl[i] = i - 1;
    while ((fl[i]) && (g[i] >= g[fl[i]])) fl[i] = fl[fl[i]];
  }
  for (int i = n - 1; i >= 1; i--) {
    fr[i] = i + 1;
    while ((fr[i] < n) && (g[i] > g[fr[i]])) fr[i] = fr[fr[i]];
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long an = 0;
    for (int i = l; i < r; i++) {
      int ll = max(l, fl[i] + 1);
      int rr = min(r - 1, fr[i] - 1);
      an += 1LL * (i - ll + 1) * (rr - i + 1) * g[i];
    }
    printf("%I64d\n", an);
  }
}

### Prompt

Please formulate a CPP solution to 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 i, j, k, l, n, m, s, an, K, X, tot = 1, h, oo, sum;
const int N = 210000;
int a[N], fl[N], fr[N], g[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), g[i - 1] = abs(a[i] - a[i - 1]);
  for (int i = l; i < n; i++) {
    fl[i] = i - 1;
    while ((fl[i]) && (g[i] >= g[fl[i]])) fl[i] = fl[fl[i]];
  }
  for (int i = n - 1; i >= 1; i--) {
    fr[i] = i + 1;
    while ((fr[i] < n) && (g[i] > g[fr[i]])) fr[i] = fr[fr[i]];
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long an = 0;
    for (int i = l; i < r; i++) {
      int ll = max(l, fl[i] + 1);
      int rr = min(r - 1, fr[i] - 1);
      an += 1LL * (i - ll + 1) * (rr - i + 1) * g[i];
    }
    printf("%I64d\n", an);
  }
}

```