#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; (i) < (n); ++(i)) cin >> a[i];
  vector<int> b(n - 1);
  for (int i = 0; (i) < (n - 1); ++(i)) b[i] = abs(a[i + 1] - a[i]);
  for (int qi = 0; (qi) < (q); ++(qi)) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    long long result = 0;
    long long acc = 0;
    map<int, int> f;
    for (int i = (l); (i) < (r); ++(i)) {
      int cnt = 0;
      for (auto it = f.begin(); it != f.end();) {
        if (b[i] <= it->first) break;
        cnt += it->second;
        acc += (((long long)b[i]) - it->first) * it->second;
        it = f.erase(it);
      }
      acc += b[i];
      f[b[i]] += cnt + 1;
      result += acc;
    }
    cout << result << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; (i) < (n); ++(i)) cin >> a[i];
  vector<int> b(n - 1);
  for (int i = 0; (i) < (n - 1); ++(i)) b[i] = abs(a[i + 1] - a[i]);
  for (int qi = 0; (qi) < (q); ++(qi)) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    long long result = 0;
    long long acc = 0;
    map<int, int> f;
    for (int i = (l); (i) < (r); ++(i)) {
      int cnt = 0;
      for (auto it = f.begin(); it != f.end();) {
        if (b[i] <= it->first) break;
        cnt += it->second;
        acc += (((long long)b[i]) - it->first) * it->second;
        it = f.erase(it);
      }
      acc += b[i];
      f[b[i]] += cnt + 1;
      result += acc;
    }
    cout << result << endl;
  }
  return 0;
}

```