#include <bits/stdc++.h>
using namespace std;
int t[500000], a[300000], b[300000], n, q;
unsigned long long L[200000], R[200000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) b[i] = abs(a[i + 1] - a[i]);
  n--;
  stack<int> s;
  for (int i = 0; i < n; i++)
    if (s.empty())
      s.push(i);
    else if (b[s.top()] >= b[i])
      s.push(i);
    else {
      int flag = 1;
      while (flag == 1 && !s.empty())
        if (b[s.top()] < b[i]) {
          R[s.top()] = i - 1;
          s.pop();
        } else
          flag = 0;
      s.push(i);
    }
  while (!s.empty()) {
    R[s.top()] = n - 1;
    s.pop();
  }
  for (int i = n - 1; i >= 0; i--)
    if (s.empty())
      s.push(i);
    else if (b[s.top()] > b[i])
      s.push(i);
    else {
      int flag = 1;
      while (flag == 1 && !s.empty())
        if (b[s.top()] <= b[i]) {
          L[s.top()] = i + 1;
          s.pop();
        } else
          flag = 0;
      s.push(i);
    }
  while (!s.empty()) {
    L[s.top()] = 0;
    s.pop();
  }
  for (int i = 0; i < q; i++) {
    unsigned long long l, r;
    cin >> l >> r;
    l--;
    r -= 2;
    unsigned long long ans = 0;
    for (unsigned long long j = l; j <= r; j++)
      ans += (unsigned long long)b[j] * (j - max(L[j], l) + 1) *
             (min(R[j], r) + 1 - j);
    cout << ans << endl;
  }
}

### Prompt

Your task is to create 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 t[500000], a[300000], b[300000], n, q;
unsigned long long L[200000], R[200000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) b[i] = abs(a[i + 1] - a[i]);
  n--;
  stack<int> s;
  for (int i = 0; i < n; i++)
    if (s.empty())
      s.push(i);
    else if (b[s.top()] >= b[i])
      s.push(i);
    else {
      int flag = 1;
      while (flag == 1 && !s.empty())
        if (b[s.top()] < b[i]) {
          R[s.top()] = i - 1;
          s.pop();
        } else
          flag = 0;
      s.push(i);
    }
  while (!s.empty()) {
    R[s.top()] = n - 1;
    s.pop();
  }
  for (int i = n - 1; i >= 0; i--)
    if (s.empty())
      s.push(i);
    else if (b[s.top()] > b[i])
      s.push(i);
    else {
      int flag = 1;
      while (flag == 1 && !s.empty())
        if (b[s.top()] <= b[i]) {
          L[s.top()] = i + 1;
          s.pop();
        } else
          flag = 0;
      s.push(i);
    }
  while (!s.empty()) {
    L[s.top()] = 0;
    s.pop();
  }
  for (int i = 0; i < q; i++) {
    unsigned long long l, r;
    cin >> l >> r;
    l--;
    r -= 2;
    unsigned long long ans = 0;
    for (unsigned long long j = l; j <= r; j++)
      ans += (unsigned long long)b[j] * (j - max(L[j], l) + 1) *
             (min(R[j], r) + 1 - j);
    cout << ans << endl;
  }
}

```