#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(0);
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  --n;
  for (int i = 0; i < (int)(n); ++i) {
    a[i] = abs(a[i + 1] - a[i]);
  }
  int cntLeft[n], cntRight[n];
  for (int turn = 0; turn < (int)(2); ++turn) {
    int prev = -1;
    int j = 0;
    int delta = turn;
    vector<int> stack;
    for (int i = 0; i < (int)(n); ++i) {
      int x = a[i];
      if (x - prev < delta) {
        stack.clear();
        j = i;
      } else {
        while (j - 1 >= 0 && x - a[j - 1] >= delta) {
          stack.push_back(a[j]);
          --j;
        }
        while (!stack.empty() && x - stack.back() < delta) {
          stack.pop_back();
          ++j;
        }
      }
      (turn ? cntRight[n - 1 - i] : cntLeft[i]) = i - j;
      prev = x;
    }
    reverse(a, a + n);
  }
  for (int t = 0; t < (int)(q); ++t) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    long long res = 0;
    for (int i = l; i <= (int)(r - 1); ++i) {
      int left = min(i - l, cntLeft[i]);
      int right = min(r - 1 - i, cntRight[i]);
      res += 1LL * a[i] * (left + 1) * (right + 1);
    }
    printf("%I64d\n", res);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(0);
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  --n;
  for (int i = 0; i < (int)(n); ++i) {
    a[i] = abs(a[i + 1] - a[i]);
  }
  int cntLeft[n], cntRight[n];
  for (int turn = 0; turn < (int)(2); ++turn) {
    int prev = -1;
    int j = 0;
    int delta = turn;
    vector<int> stack;
    for (int i = 0; i < (int)(n); ++i) {
      int x = a[i];
      if (x - prev < delta) {
        stack.clear();
        j = i;
      } else {
        while (j - 1 >= 0 && x - a[j - 1] >= delta) {
          stack.push_back(a[j]);
          --j;
        }
        while (!stack.empty() && x - stack.back() < delta) {
          stack.pop_back();
          ++j;
        }
      }
      (turn ? cntRight[n - 1 - i] : cntLeft[i]) = i - j;
      prev = x;
    }
    reverse(a, a + n);
  }
  for (int t = 0; t < (int)(q); ++t) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    long long res = 0;
    for (int i = l; i <= (int)(r - 1); ++i) {
      int left = min(i - l, cntLeft[i]);
      int right = min(r - 1 - i, cntRight[i]);
      res += 1LL * a[i] * (left + 1) * (right + 1);
    }
    printf("%I64d\n", res);
  }
  return 0;
}

```