#include <bits/stdc++.h>
using namespace std;
long long a[1000000], l[1000000], r[1000000], diff[1000000];
int main() {
  long long n, q, i, j, p, x, y, c, d;
  scanf("%lld %lld", &n, &q);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 1; i < n; i++) {
    a[i] = abs(a[i + 1] - a[i]);
  }
  stack<long long> s;
  for (i = 1; i <= n - 1; i++) {
    if (s.empty()) {
      s.push(i);
    } else {
      p = s.top();
      if (a[p] > a[i]) {
        s.push(i);
      } else {
        while (a[s.top()] <= a[i]) {
          p = s.top();
          s.pop();
          l[p] = i - 1;
          if (s.empty()) break;
        }
        s.push(i);
      }
    }
  }
  while (!s.empty()) {
    p = s.top();
    l[p] = n - 1;
    s.pop();
  }
  for (i = n - 1; i >= 0; i--) {
    if (s.empty()) {
      s.push(i);
    } else {
      p = s.top();
      if (a[p] > a[i]) {
        s.push(i);
      } else {
        while (a[s.top()] < a[i]) {
          p = s.top();
          s.pop();
          r[p] = i + 1;
          if (s.empty()) break;
        }
        s.push(i);
      }
    }
  }
  while (!s.empty()) {
    p = s.top();
    r[p] = 1;
    s.pop();
  }
  while (q--) {
    long long ans = 0;
    scanf("%lld %lld", &x, &y);
    for (i = x; i < y; i++) {
      c = i - max(x, r[i]);
      d = min(y - 1, l[i]) - i;
      ans += (c * d + c + d + 1) * a[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long a[1000000], l[1000000], r[1000000], diff[1000000];
int main() {
  long long n, q, i, j, p, x, y, c, d;
  scanf("%lld %lld", &n, &q);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 1; i < n; i++) {
    a[i] = abs(a[i + 1] - a[i]);
  }
  stack<long long> s;
  for (i = 1; i <= n - 1; i++) {
    if (s.empty()) {
      s.push(i);
    } else {
      p = s.top();
      if (a[p] > a[i]) {
        s.push(i);
      } else {
        while (a[s.top()] <= a[i]) {
          p = s.top();
          s.pop();
          l[p] = i - 1;
          if (s.empty()) break;
        }
        s.push(i);
      }
    }
  }
  while (!s.empty()) {
    p = s.top();
    l[p] = n - 1;
    s.pop();
  }
  for (i = n - 1; i >= 0; i--) {
    if (s.empty()) {
      s.push(i);
    } else {
      p = s.top();
      if (a[p] > a[i]) {
        s.push(i);
      } else {
        while (a[s.top()] < a[i]) {
          p = s.top();
          s.pop();
          r[p] = i + 1;
          if (s.empty()) break;
        }
        s.push(i);
      }
    }
  }
  while (!s.empty()) {
    p = s.top();
    r[p] = 1;
    s.pop();
  }
  while (q--) {
    long long ans = 0;
    scanf("%lld %lld", &x, &y);
    for (i = x; i < y; i++) {
      c = i - max(x, r[i]);
      d = min(y - 1, l[i]) - i;
      ans += (c * d + c + d + 1) * a[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```