#include <bits/stdc++.h>
using namespace std;
int n, Q, inp[101000], w[101000];
int p[110][2];
long long Res[110];
struct point {
  int g, x;
  bool operator<(const point &p) const { return g < p.g; }
} ord[101000];
set<int> Set;
int main() {
  int i, b, e, j, b2, e2;
  set<int>::iterator it, it2;
  scanf("%d%d", &n, &Q);
  for (i = 1; i <= n; i++) scanf("%d", &inp[i]);
  for (i = 1; i < n; i++) w[i] = abs(inp[i + 1] - inp[i]);
  for (i = 1; i <= Q; i++) {
    scanf("%d%d", &p[i][0], &p[i][1]);
    p[i][1]--;
  }
  Set.insert(0);
  Set.insert(n);
  for (i = 1; i < n; i++) {
    ord[i].x = i;
    ord[i].g = w[i];
  }
  sort(ord + 1, ord + n);
  for (i = n - 1; i >= 1; i--) {
    it = Set.lower_bound(ord[i].x);
    it2 = it;
    it--;
    b = (*it) + 1;
    e = (*it2) - 1;
    for (j = 1; j <= Q; j++) {
      b2 = max(b, p[j][0]);
      e2 = min(e, p[j][1]);
      if (b2 <= ord[i].x && ord[i].x <= e2) {
        Res[j] +=
            (long long)ord[i].g * (ord[i].x - b2 + 1) * (e2 - ord[i].x + 1);
      }
    }
    Set.insert(ord[i].x);
  }
  for (i = 1; i <= Q; i++) printf("%lld\n", Res[i]);
}

### Prompt

Generate 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 n, Q, inp[101000], w[101000];
int p[110][2];
long long Res[110];
struct point {
  int g, x;
  bool operator<(const point &p) const { return g < p.g; }
} ord[101000];
set<int> Set;
int main() {
  int i, b, e, j, b2, e2;
  set<int>::iterator it, it2;
  scanf("%d%d", &n, &Q);
  for (i = 1; i <= n; i++) scanf("%d", &inp[i]);
  for (i = 1; i < n; i++) w[i] = abs(inp[i + 1] - inp[i]);
  for (i = 1; i <= Q; i++) {
    scanf("%d%d", &p[i][0], &p[i][1]);
    p[i][1]--;
  }
  Set.insert(0);
  Set.insert(n);
  for (i = 1; i < n; i++) {
    ord[i].x = i;
    ord[i].g = w[i];
  }
  sort(ord + 1, ord + n);
  for (i = n - 1; i >= 1; i--) {
    it = Set.lower_bound(ord[i].x);
    it2 = it;
    it--;
    b = (*it) + 1;
    e = (*it2) - 1;
    for (j = 1; j <= Q; j++) {
      b2 = max(b, p[j][0]);
      e2 = min(e, p[j][1]);
      if (b2 <= ord[i].x && ord[i].x <= e2) {
        Res[j] +=
            (long long)ord[i].g * (ord[i].x - b2 + 1) * (e2 - ord[i].x + 1);
      }
    }
    Set.insert(ord[i].x);
  }
  for (i = 1; i <= Q; i++) printf("%lld\n", Res[i]);
}

```