#include <bits/stdc++.h>
using namespace std;
int n, q;
vector<int> A;
vector<pair<int, int>> Q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    A.push_back(x);
  }
  for (int i = 0; i < q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    Q.push_back(make_pair(a - 1, b - 1));
  }
  priority_queue<pair<int, int>> que;
  for (int i = 1; i < n; ++i) {
    que.push(make_pair(abs(A[i - 1] - A[i]), i - 1));
  }
  set<int> q1, q2;
  q1.insert(-1);
  q1.insert(n);
  q2.insert(-1);
  q2.insert(n);
  --n;
  vector<long long> res(q, 0);
  while (que.size()) {
    int val = que.top().first;
    int pos = que.top().second;
    que.pop();
    int a = n - *q1.upper_bound(n - pos);
    int b = *q2.upper_bound(pos);
    for (int i = 0; i < q; ++i) {
      int L = Q[i].first - 1, R = Q[i].second;
      if (pos > L && pos < R) {
        res[i] += (pos - max(L, a)) * (long long)(min(R, b) - pos) * val;
      }
    }
    q1.insert(n - pos);
    q2.insert(pos);
  }
  for (auto x : res) {
    printf("%I64d\n", x);
  }
  return 0;
}

### Prompt

Please formulate 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;
vector<int> A;
vector<pair<int, int>> Q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    A.push_back(x);
  }
  for (int i = 0; i < q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    Q.push_back(make_pair(a - 1, b - 1));
  }
  priority_queue<pair<int, int>> que;
  for (int i = 1; i < n; ++i) {
    que.push(make_pair(abs(A[i - 1] - A[i]), i - 1));
  }
  set<int> q1, q2;
  q1.insert(-1);
  q1.insert(n);
  q2.insert(-1);
  q2.insert(n);
  --n;
  vector<long long> res(q, 0);
  while (que.size()) {
    int val = que.top().first;
    int pos = que.top().second;
    que.pop();
    int a = n - *q1.upper_bound(n - pos);
    int b = *q2.upper_bound(pos);
    for (int i = 0; i < q; ++i) {
      int L = Q[i].first - 1, R = Q[i].second;
      if (pos > L && pos < R) {
        res[i] += (pos - max(L, a)) * (long long)(min(R, b) - pos) * val;
      }
    }
    q1.insert(n - pos);
    q2.insert(pos);
  }
  for (auto x : res) {
    printf("%I64d\n", x);
  }
  return 0;
}

```