#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, q;
  cin >> n >> q;
  vector<long long int> a(n);
  for (long long int i = 0; i != n; i++) {
    cin >> a[i];
  }
  vector<long long int> h(n - 1);
  for (long long int i = 0; i != n - 1; i++) {
    h[i] = abs(a[i] - a[i + 1]);
  }
  vector<long long int> left(n - 1, -1);
  vector<long long int> right(n - 1, n - 1);
  stack<long long int> s;
  stack<long long int> t;
  t.push(0);
  s.push(n - 2);
  for (long long int i = 1; i != n - 1; i++) {
    if (h[i] > h[t.top()]) {
      while (t.size() != 0 && h[i] > h[t.top()]) {
        right[t.top()] = i;
        t.pop();
      }
      t.push(i);
    } else {
      t.push(i);
    }
  }
  for (long long int i = n - 3; i != -1; i--) {
    if (h[i] >= h[s.top()]) {
      while (s.size() != 0 && h[i] >= h[s.top()]) {
        left[s.top()] = i;
        s.pop();
      }
      s.push(i);
    } else {
      s.push(i);
    }
  }
  for (long long int i = 0; i != n - 1; i++) {
    right[i]--;
    left[i]++;
  }
  long long int l, r;
  for (long long int i = 0; i != q; i++) {
    cin >> l >> r;
    l--;
    r--;
    long long int x, y, ans = 0;
    for (long long int j = l; j != r; j++) {
      x = max(left[j], l);
      y = min(right[j], r - 1);
      ans += (y - j + 1) * (j - x + 1) * h[j];
    }
    cout << ans << endl;
  }
}

### 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;
int main() {
  long long int n, q;
  cin >> n >> q;
  vector<long long int> a(n);
  for (long long int i = 0; i != n; i++) {
    cin >> a[i];
  }
  vector<long long int> h(n - 1);
  for (long long int i = 0; i != n - 1; i++) {
    h[i] = abs(a[i] - a[i + 1]);
  }
  vector<long long int> left(n - 1, -1);
  vector<long long int> right(n - 1, n - 1);
  stack<long long int> s;
  stack<long long int> t;
  t.push(0);
  s.push(n - 2);
  for (long long int i = 1; i != n - 1; i++) {
    if (h[i] > h[t.top()]) {
      while (t.size() != 0 && h[i] > h[t.top()]) {
        right[t.top()] = i;
        t.pop();
      }
      t.push(i);
    } else {
      t.push(i);
    }
  }
  for (long long int i = n - 3; i != -1; i--) {
    if (h[i] >= h[s.top()]) {
      while (s.size() != 0 && h[i] >= h[s.top()]) {
        left[s.top()] = i;
        s.pop();
      }
      s.push(i);
    } else {
      s.push(i);
    }
  }
  for (long long int i = 0; i != n - 1; i++) {
    right[i]--;
    left[i]++;
  }
  long long int l, r;
  for (long long int i = 0; i != q; i++) {
    cin >> l >> r;
    l--;
    r--;
    long long int x, y, ans = 0;
    for (long long int j = l; j != r; j++) {
      x = max(left[j], l);
      y = min(right[j], r - 1);
      ans += (y - j + 1) * (j - x + 1) * h[j];
    }
    cout << ans << endl;
  }
}

```