#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const long long inf = 1e17 + 10;
using namespace std;
int n, q;
int peso[MAXN][110];
long long a[MAXN], curDir[MAXN];
int main() {
  for (int i = 0; i < MAXN; i++)
    for (int j = 0; j < 110; j++) peso[i][j] = 1;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = n - 1; i > 0; i--) a[i] = abs(a[i] - a[i - 1]);
  for (int i = 0; i < q; i++) {
    int l, r;
    long long resp = 0;
    scanf("%d%d", &l, &r);
    r--;
    long long aux1 = a[r + 1];
    a[r + 1] = inf;
    for (int g = r; g >= l; g--) {
      int cur = g + 1;
      while (a[cur] <= a[g]) cur = curDir[cur];
      curDir[g] = cur;
    }
    a[r + 1] = aux1;
    for (int g = l; g <= r; g++) {
      resp += a[g] * (curDir[g] - g) * peso[g][i];
      peso[curDir[g]][i] += peso[g][i];
    }
    printf("%lld\n", resp);
  }
}

### 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>
const int MAXN = 1e5 + 10;
const long long inf = 1e17 + 10;
using namespace std;
int n, q;
int peso[MAXN][110];
long long a[MAXN], curDir[MAXN];
int main() {
  for (int i = 0; i < MAXN; i++)
    for (int j = 0; j < 110; j++) peso[i][j] = 1;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = n - 1; i > 0; i--) a[i] = abs(a[i] - a[i - 1]);
  for (int i = 0; i < q; i++) {
    int l, r;
    long long resp = 0;
    scanf("%d%d", &l, &r);
    r--;
    long long aux1 = a[r + 1];
    a[r + 1] = inf;
    for (int g = r; g >= l; g--) {
      int cur = g + 1;
      while (a[cur] <= a[g]) cur = curDir[cur];
      curDir[g] = cur;
    }
    a[r + 1] = aux1;
    for (int g = l; g <= r; g++) {
      resp += a[g] * (curDir[g] - g) * peso[g][i];
      peso[curDir[g]][i] += peso[g][i];
    }
    printf("%lld\n", resp);
  }
}

```