#include <bits/stdc++.h>
using namespace std;
long long a[1000006];
long long b[1000006], c[1000006];
stack<int> my;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q, i;
  cin >> n >> q;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = n; i > 1; i--) a[i] = abs(a[i] - a[i - 1]);
  my.push(2);
  i = 3;
  while (i <= n) {
    while (i <= n && (my.size() > 0) && a[i] <= a[my.top()]) {
      my.push(i);
      i++;
    }
    while (i <= n && (my.size() > 0) && a[i] > a[my.top()]) {
      int j = my.top();
      c[j] = i - j - 1;
      my.pop();
    }
    if (i <= n) {
      if (my.size() > 0)
        b[i] = i - my.top() - 1;
      else
        b[i] = i - 1;
      my.push(i);
      i++;
    }
  }
  while (my.size() > 0) {
    c[my.top()] = n - my.top();
    my.pop();
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    long long ans = 0;
    for (i = l + 1; i <= r; i++)
      ans += (min(b[i] + 1, (long long)i - l) *
              min(c[i] + 1, (long long)r - i + 1)) *
             a[i];
    cout << ans << endl;
  }
  return 0;
}

### 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;
long long a[1000006];
long long b[1000006], c[1000006];
stack<int> my;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q, i;
  cin >> n >> q;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = n; i > 1; i--) a[i] = abs(a[i] - a[i - 1]);
  my.push(2);
  i = 3;
  while (i <= n) {
    while (i <= n && (my.size() > 0) && a[i] <= a[my.top()]) {
      my.push(i);
      i++;
    }
    while (i <= n && (my.size() > 0) && a[i] > a[my.top()]) {
      int j = my.top();
      c[j] = i - j - 1;
      my.pop();
    }
    if (i <= n) {
      if (my.size() > 0)
        b[i] = i - my.top() - 1;
      else
        b[i] = i - 1;
      my.push(i);
      i++;
    }
  }
  while (my.size() > 0) {
    c[my.top()] = n - my.top();
    my.pop();
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    long long ans = 0;
    for (i = l + 1; i <= r; i++)
      ans += (min(b[i] + 1, (long long)i - l) *
              min(c[i] + 1, (long long)r - i + 1)) *
             a[i];
    cout << ans << endl;
  }
  return 0;
}

```