#include <bits/stdc++.h>
using namespace std;
int w[110000], a[110000], n, x[110000], f[110000], size[110000];
int compare(int k1, int k2) { return w[k1] < w[k2]; }
int findfather(int k1) {
  if (f[k1] == k1) return k1;
  f[k1] = findfather(f[k1]);
  return f[k1];
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) w[i] = abs(x[i] - x[i - 1]), a[i] = i;
  sort(a + 1, a + n + 1, compare);
  for (; q; q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int i = 1; i <= n; i++) f[i] = i, size[i] = 1;
    long long ans = 0;
    for (int now = 1; now <= n; now++)
      if (a[now] > l && a[now] <= r) {
        int k1 = findfather(a[now] - 1), k2 = findfather(a[now]);
        ans = (ans + 1ll * size[k1] * size[k2] * w[a[now]]);
        if (size[k1] > size[k2]) swap(k1, k2);
        f[k1] = k2;
        size[k2] += size[k1];
      }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 w[110000], a[110000], n, x[110000], f[110000], size[110000];
int compare(int k1, int k2) { return w[k1] < w[k2]; }
int findfather(int k1) {
  if (f[k1] == k1) return k1;
  f[k1] = findfather(f[k1]);
  return f[k1];
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) w[i] = abs(x[i] - x[i - 1]), a[i] = i;
  sort(a + 1, a + n + 1, compare);
  for (; q; q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int i = 1; i <= n; i++) f[i] = i, size[i] = 1;
    long long ans = 0;
    for (int now = 1; now <= n; now++)
      if (a[now] > l && a[now] <= r) {
        int k1 = findfather(a[now] - 1), k2 = findfather(a[now]);
        ans = (ans + 1ll * size[k1] * size[k2] * w[a[now]]);
        if (size[k1] > size[k2]) swap(k1, k2);
        f[k1] = k2;
        size[k2] += size[k1];
      }
    printf("%I64d\n", ans);
  }
  return 0;
}

```