#include <bits/stdc++.h>
using namespace std;
int n, q;
int s[100001];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
  }
  for (int ll = 0; ll < q; ll++) {
    int k, v;
    scanf("%d%d", &k, &v);
    long long sum = 0;
    vector<pair<long long, int>> kul;
    vector<int> hol;
    for (int i = k; i < v; i++) {
      while (kul.size() > 0 &&
             (kul[kul.size() - 1].first <= abs(s[i + 1] - s[i]))) {
        sum += (long long)kul[kul.size() - 1].first *
               (kul[kul.size() - 1].second) * (i - hol[kul.size() - 1]);
        kul.erase(--kul.end(), kul.end());
        hol.erase(--hol.end(), hol.end());
      }
      if (kul.size() > 0) {
        kul.push_back(make_pair(abs(s[i + 1] - s[i]), i - hol[kul.size() - 1]));
      } else {
        kul.push_back(make_pair(abs(s[i + 1] - s[i]), i - k + 1));
      }
      hol.push_back(i);
    }
    while (kul.size() > 0) {
      sum += (long long)kul[kul.size() - 1].first *
             (kul[kul.size() - 1].second) * (v - hol[kul.size() - 1]);
      kul.erase(--kul.end(), kul.end());
      hol.erase(--hol.end(), hol.end());
    }
    printf("%I64d\n", sum);
  }
  return 0;
}

### 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 n, q;
int s[100001];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
  }
  for (int ll = 0; ll < q; ll++) {
    int k, v;
    scanf("%d%d", &k, &v);
    long long sum = 0;
    vector<pair<long long, int>> kul;
    vector<int> hol;
    for (int i = k; i < v; i++) {
      while (kul.size() > 0 &&
             (kul[kul.size() - 1].first <= abs(s[i + 1] - s[i]))) {
        sum += (long long)kul[kul.size() - 1].first *
               (kul[kul.size() - 1].second) * (i - hol[kul.size() - 1]);
        kul.erase(--kul.end(), kul.end());
        hol.erase(--hol.end(), hol.end());
      }
      if (kul.size() > 0) {
        kul.push_back(make_pair(abs(s[i + 1] - s[i]), i - hol[kul.size() - 1]));
      } else {
        kul.push_back(make_pair(abs(s[i + 1] - s[i]), i - k + 1));
      }
      hol.push_back(i);
    }
    while (kul.size() > 0) {
      sum += (long long)kul[kul.size() - 1].first *
             (kul[kul.size() - 1].second) * (v - hol[kul.size() - 1]);
      kul.erase(--kul.end(), kul.end());
      hol.erase(--hol.end(), hol.end());
    }
    printf("%I64d\n", sum);
  }
  return 0;
}

```