#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct elem {
  int num, id, l, r;
} a[maxn];
int num[maxn];
int main() {
  int n, q;
  while (scanf("%d%d", &n, &q) != EOF) {
    for (int i = 1; i <= n; ++i) scanf("%d", &num[i]);
    for (int i = 1; i < n; ++i) {
      int tmp = num[i + 1] - num[i];
      a[i].num = tmp >= 0 ? tmp : -tmp;
      a[i].id = i;
      a[i].l = 0;
      a[i].r = n;
    }
    stack<elem> s;
    for (int i = 1; i < n; ++i) {
      while (!s.empty()) {
        elem u = s.top();
        if (u.num < a[i].num) {
          a[u.id].r = a[i].id;
          s.pop();
        } else
          break;
      }
      s.push(a[i]);
    }
    while (!s.empty()) s.pop();
    for (int i = n - 1; i >= 1; --i) {
      while (!s.empty()) {
        elem u = s.top();
        if (u.num <= a[i].num) {
          a[u.id].l = a[i].id;
          s.pop();
        } else
          break;
      }
      s.push(a[i]);
    }
    while (q--) {
      long long ans = 0;
      int l, r;
      scanf("%d%d", &l, &r);
      for (int i = l; i < r; ++i) {
        int num1, num2;
        if (a[i].l < l - 1)
          num1 = i - (l - 1);
        else
          num1 = i - a[i].l;
        if (a[i].r > r)
          num2 = r - i;
        else
          num2 = a[i].r - i;
        ans += a[i].num * (long long)num1 * num2;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct elem {
  int num, id, l, r;
} a[maxn];
int num[maxn];
int main() {
  int n, q;
  while (scanf("%d%d", &n, &q) != EOF) {
    for (int i = 1; i <= n; ++i) scanf("%d", &num[i]);
    for (int i = 1; i < n; ++i) {
      int tmp = num[i + 1] - num[i];
      a[i].num = tmp >= 0 ? tmp : -tmp;
      a[i].id = i;
      a[i].l = 0;
      a[i].r = n;
    }
    stack<elem> s;
    for (int i = 1; i < n; ++i) {
      while (!s.empty()) {
        elem u = s.top();
        if (u.num < a[i].num) {
          a[u.id].r = a[i].id;
          s.pop();
        } else
          break;
      }
      s.push(a[i]);
    }
    while (!s.empty()) s.pop();
    for (int i = n - 1; i >= 1; --i) {
      while (!s.empty()) {
        elem u = s.top();
        if (u.num <= a[i].num) {
          a[u.id].l = a[i].id;
          s.pop();
        } else
          break;
      }
      s.push(a[i]);
    }
    while (q--) {
      long long ans = 0;
      int l, r;
      scanf("%d%d", &l, &r);
      for (int i = l; i < r; ++i) {
        int num1, num2;
        if (a[i].l < l - 1)
          num1 = i - (l - 1);
        else
          num1 = i - a[i].l;
        if (a[i].r > r)
          num2 = r - i;
        else
          num2 = a[i].r - i;
        ans += a[i].num * (long long)num1 * num2;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```