#include <bits/stdc++.h>
int main() {
  int n, q;
  std::cin >> n >> q;
  std::vector<int> numbers(n);
  std::vector<long long> differences(n - 1);
  for (int i = 0; i < n; ++i) {
    std::cin >> numbers[i];
    if (i > 0) differences[i - 1] = std::abs(numbers[i] - numbers[i - 1]);
  }
  std::vector<int> left(n - 1), right(n - 1);
  std::stack<std::pair<int, int>> previous;
  for (int i = 0; i < n - 1; ++i) {
    while (!previous.empty() && previous.top().first < differences[i]) {
      previous.pop();
    }
    left[i] = (previous.empty() ? -1 : previous.top().second);
    previous.push(std::make_pair(differences[i], i));
  }
  while (!previous.empty()) previous.pop();
  for (int i = n - 2; i >= 0; --i) {
    while (!previous.empty() && previous.top().first <= differences[i]) {
      previous.pop();
    }
    right[i] = (previous.empty() ? n : previous.top().second);
    previous.push(std::make_pair(differences[i], i));
  }
  for (int query_index = 0; query_index < q; ++query_index) {
    int l, r;
    std::cin >> l >> r;
    l--, r--;
    long long answer = 0;
    for (int i = l; i < r; ++i) {
      answer += differences[i] * (long long)std::min(i - l + 1, i - left[i]) *
                (long long)std::min(r - i, right[i] - i);
    }
    std::cout << answer << std::endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
int main() {
  int n, q;
  std::cin >> n >> q;
  std::vector<int> numbers(n);
  std::vector<long long> differences(n - 1);
  for (int i = 0; i < n; ++i) {
    std::cin >> numbers[i];
    if (i > 0) differences[i - 1] = std::abs(numbers[i] - numbers[i - 1]);
  }
  std::vector<int> left(n - 1), right(n - 1);
  std::stack<std::pair<int, int>> previous;
  for (int i = 0; i < n - 1; ++i) {
    while (!previous.empty() && previous.top().first < differences[i]) {
      previous.pop();
    }
    left[i] = (previous.empty() ? -1 : previous.top().second);
    previous.push(std::make_pair(differences[i], i));
  }
  while (!previous.empty()) previous.pop();
  for (int i = n - 2; i >= 0; --i) {
    while (!previous.empty() && previous.top().first <= differences[i]) {
      previous.pop();
    }
    right[i] = (previous.empty() ? n : previous.top().second);
    previous.push(std::make_pair(differences[i], i));
  }
  for (int query_index = 0; query_index < q; ++query_index) {
    int l, r;
    std::cin >> l >> r;
    l--, r--;
    long long answer = 0;
    for (int i = l; i < r; ++i) {
      answer += differences[i] * (long long)std::min(i - l + 1, i - left[i]) *
                (long long)std::min(r - i, right[i] - i);
    }
    std::cout << answer << std::endl;
  }
  return 0;
}

```