#include <bits/stdc++.h>
using namespace std;
long long a[100500];
long long n, q;
long long sparse[100500][20];
void proc(long long* a, int n) {
  for (int i = 0; i < n; i++) {
    sparse[i][0] = i;
  }
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 0; i + (1 << j) - 1 < n; i++) {
      if (a[sparse[i][j - 1]] > a[sparse[i + (1 << (j - 1))][j - 1]]) {
        sparse[i][j] = sparse[i][j - 1];
      } else {
        sparse[i][j] = sparse[i + (1 << (j - 1))][j - 1];
      }
    }
  }
}
long long max(int l, int r) {
  int k = r - l + 1;
  int ll = log(k) / log(2);
  if (a[sparse[l][ll]] > a[sparse[l + k - (1 << ll)][ll]]) {
    return sparse[l][ll];
  } else {
    return sparse[l + k - (1 << ll)][ll];
  }
}
long long rez(int l, int r) {
  if (l == r) return a[l];
  if (l > r) return 0;
  long long rezult = 0;
  long long pos = max(l, r);
  rezult = 1LL * a[pos] * (r - pos + 1) * (pos - l + 1);
  rezult += rez(l, pos - 1) + rez(pos + 1, r);
  return rezult;
}
int main() {
  long long k = 0;
  long long prev = 0;
  cin >> n;
  cin >> q;
  a[0] = 0;
  cin >> prev;
  for (int i = 1; i < n; i++) {
    cin >> k;
    a[i] = abs(prev - k);
    prev = k;
  }
  proc(a, n);
  int x = 0, y = 0;
  for (int i = 0; i < q; i++) {
    cin >> x >> y;
    cout << rez(x, y - 1) << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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[100500];
long long n, q;
long long sparse[100500][20];
void proc(long long* a, int n) {
  for (int i = 0; i < n; i++) {
    sparse[i][0] = i;
  }
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 0; i + (1 << j) - 1 < n; i++) {
      if (a[sparse[i][j - 1]] > a[sparse[i + (1 << (j - 1))][j - 1]]) {
        sparse[i][j] = sparse[i][j - 1];
      } else {
        sparse[i][j] = sparse[i + (1 << (j - 1))][j - 1];
      }
    }
  }
}
long long max(int l, int r) {
  int k = r - l + 1;
  int ll = log(k) / log(2);
  if (a[sparse[l][ll]] > a[sparse[l + k - (1 << ll)][ll]]) {
    return sparse[l][ll];
  } else {
    return sparse[l + k - (1 << ll)][ll];
  }
}
long long rez(int l, int r) {
  if (l == r) return a[l];
  if (l > r) return 0;
  long long rezult = 0;
  long long pos = max(l, r);
  rezult = 1LL * a[pos] * (r - pos + 1) * (pos - l + 1);
  rezult += rez(l, pos - 1) + rez(pos + 1, r);
  return rezult;
}
int main() {
  long long k = 0;
  long long prev = 0;
  cin >> n;
  cin >> q;
  a[0] = 0;
  cin >> prev;
  for (int i = 1; i < n; i++) {
    cin >> k;
    a[i] = abs(prev - k);
    prev = k;
  }
  proc(a, n);
  int x = 0, y = 0;
  for (int i = 0; i < q; i++) {
    cin >> x >> y;
    cout << rez(x, y - 1) << endl;
  }
  return 0;
}

```