#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, q;
  cin >> n >> q;
  vector<int64_t> a(n);
  vector<int64_t> D(n - 1);
  for (int64_t i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0) {
      D[i - 1] = abs(a[i - 1] - a[i]);
    }
  }
  vector<int64_t> L(n - 1);
  vector<int64_t> R(n - 1);
  stack<pair<int64_t, int64_t> > S;
  for (int64_t i = n - 2; i >= 0; i--) {
    while (!S.empty() && S.top().first <= D[i]) {
      L[S.top().second] = i + 1;
      S.pop();
    }
    S.push(make_pair(D[i], i));
  }
  while (!S.empty()) {
    L[S.top().second] = 0;
    S.pop();
  }
  for (int64_t i = 0; i < n - 1; i++) {
    while (!S.empty() && S.top().first < D[i]) {
      R[S.top().second] = i + 1;
      S.pop();
    }
    S.push(make_pair(D[i], i));
  }
  while (!S.empty()) {
    R[S.top().second] = n;
    S.pop();
  }
  for (int64_t i = 0; i < q; i++) {
    int64_t l, r;
    int64_t answer = 0;
    cin >> l >> r;
    for (int64_t j = l; j < r; j++) {
      answer += D[j - 1] * (j - max(L[j - 1], l - 1)) * (min(R[j - 1], r) - j);
    }
    cout << answer << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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<int64_t> a(n);
  vector<int64_t> D(n - 1);
  for (int64_t i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0) {
      D[i - 1] = abs(a[i - 1] - a[i]);
    }
  }
  vector<int64_t> L(n - 1);
  vector<int64_t> R(n - 1);
  stack<pair<int64_t, int64_t> > S;
  for (int64_t i = n - 2; i >= 0; i--) {
    while (!S.empty() && S.top().first <= D[i]) {
      L[S.top().second] = i + 1;
      S.pop();
    }
    S.push(make_pair(D[i], i));
  }
  while (!S.empty()) {
    L[S.top().second] = 0;
    S.pop();
  }
  for (int64_t i = 0; i < n - 1; i++) {
    while (!S.empty() && S.top().first < D[i]) {
      R[S.top().second] = i + 1;
      S.pop();
    }
    S.push(make_pair(D[i], i));
  }
  while (!S.empty()) {
    R[S.top().second] = n;
    S.pop();
  }
  for (int64_t i = 0; i < q; i++) {
    int64_t l, r;
    int64_t answer = 0;
    cin >> l >> r;
    for (int64_t j = l; j < r; j++) {
      answer += D[j - 1] * (j - max(L[j - 1], l - 1)) * (min(R[j - 1], r) - j);
    }
    cout << answer << endl;
  }
  return 0;
}

```