#include <bits/stdc++.h>
int n, q, a[100005], d[100005], l[100005], r[100005], h, b, c;
long long ans;
struct stack {
  int v, i;
} s[100005];
int min(int x, int y) { return x < y ? x : y; }
int max(int x, int y) { return x > y ? x : y; }
int abs(int x) { return x > 0 ? x : -x; }
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    l[i] = 0;
    r[i] = n + 1;
    scanf("%d", &a[i]);
    d[i] = abs(a[i] - a[i - 1]);
    while (h && d[i] > s[h - 1].v) r[s[--h].i] = i;
    if (h) l[i] = s[h - 1].i;
    s[h].v = d[i];
    s[h++].i = i;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &b, &c);
    ans = 0;
    for (int j = b; j <= c; j++)
      ans += (long long)d[j] * (j - max(l[j], b)) * (min(r[j], c + 1) - j);
    printf("%I64d\n", ans);
  }
  scanf("\n");
  return 0;
}

### Prompt

Please create a solution in cpp 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>
int n, q, a[100005], d[100005], l[100005], r[100005], h, b, c;
long long ans;
struct stack {
  int v, i;
} s[100005];
int min(int x, int y) { return x < y ? x : y; }
int max(int x, int y) { return x > y ? x : y; }
int abs(int x) { return x > 0 ? x : -x; }
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    l[i] = 0;
    r[i] = n + 1;
    scanf("%d", &a[i]);
    d[i] = abs(a[i] - a[i - 1]);
    while (h && d[i] > s[h - 1].v) r[s[--h].i] = i;
    if (h) l[i] = s[h - 1].i;
    s[h].v = d[i];
    s[h++].i = i;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &b, &c);
    ans = 0;
    for (int j = b; j <= c; j++)
      ans += (long long)d[j] * (j - max(l[j], b)) * (min(r[j], c + 1) - j);
    printf("%I64d\n", ans);
  }
  scanf("\n");
  return 0;
}

```