#include <bits/stdc++.h>
using namespace std;
long long arr[300005], a[300005];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int(i) = (0); (i) < (n); i++) scanf("%d", &arr[i]);
  for (int(i) = (0); (i) < (n - 1); i++) a[i] = abs(arr[i + 1] - arr[i]);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    stack<pair<long long, long long> > mystack;
    long long ans = 0, total = 0;
    for (int i = l; i < r; i++) {
      long long num = a[i], count = 1;
      total += num;
      while (mystack.size() > 0 and mystack.top().first <= num) {
        pair<long long, long long> top = mystack.top();
        mystack.pop();
        count += top.second;
        total -= top.first * top.second;
        total += num * top.second;
      }
      mystack.push(make_pair(num, count));
      ans += total;
    }
    cout << ans << endl;
  }
}

### 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;
long long arr[300005], a[300005];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int(i) = (0); (i) < (n); i++) scanf("%d", &arr[i]);
  for (int(i) = (0); (i) < (n - 1); i++) a[i] = abs(arr[i + 1] - arr[i]);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    stack<pair<long long, long long> > mystack;
    long long ans = 0, total = 0;
    for (int i = l; i < r; i++) {
      long long num = a[i], count = 1;
      total += num;
      while (mystack.size() > 0 and mystack.top().first <= num) {
        pair<long long, long long> top = mystack.top();
        mystack.pop();
        count += top.second;
        total -= top.first * top.second;
        total += num * top.second;
      }
      mystack.push(make_pair(num, count));
      ans += total;
    }
    cout << ans << endl;
  }
}

```