#include <bits/stdc++.h>
using namespace std;
long long a[100005], D[100005], lft[100005], rgt[100005];
long long solve(int x, int y) {
  stack<int> S;
  S.push(x - 1);
  for (int i = x; i <= y; i++) {
    while (S.top() != x - 1 && D[S.top()] <= D[i]) S.pop();
    lft[i] = S.top() + 1;
    S.push(i);
  }
  while (S.size() > 0) S.pop();
  S.push(y + 1);
  for (int i = y; i >= x; i--) {
    while (S.top() != y + 1 && D[S.top()] < D[i]) S.pop();
    rgt[i] = S.top() - 1;
    S.push(i);
  }
  long long ans = 0;
  for (int i = x; i <= y; i++) {
    ans += (long long)D[i] * (long long)(i - lft[i] + 1) *
           (long long)(rgt[i] - i + 1);
  }
  return ans;
}
int main(void) {
  long n, q, temp, maxlen = -1;
  cin >> n;
  cin >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) D[i] = abs(a[i] - a[i + 1]);
  int l, r;
  while (q--) {
    long ans = 0;
    cin >> l >> r;
    l--;
    r--;
    cout << solve(l, r - 1) << '\n';
  }
  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;
long long a[100005], D[100005], lft[100005], rgt[100005];
long long solve(int x, int y) {
  stack<int> S;
  S.push(x - 1);
  for (int i = x; i <= y; i++) {
    while (S.top() != x - 1 && D[S.top()] <= D[i]) S.pop();
    lft[i] = S.top() + 1;
    S.push(i);
  }
  while (S.size() > 0) S.pop();
  S.push(y + 1);
  for (int i = y; i >= x; i--) {
    while (S.top() != y + 1 && D[S.top()] < D[i]) S.pop();
    rgt[i] = S.top() - 1;
    S.push(i);
  }
  long long ans = 0;
  for (int i = x; i <= y; i++) {
    ans += (long long)D[i] * (long long)(i - lft[i] + 1) *
           (long long)(rgt[i] - i + 1);
  }
  return ans;
}
int main(void) {
  long n, q, temp, maxlen = -1;
  cin >> n;
  cin >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) D[i] = abs(a[i] - a[i + 1]);
  int l, r;
  while (q--) {
    long ans = 0;
    cin >> l >> r;
    l--;
    r--;
    cout << solve(l, r - 1) << '\n';
  }
  return 0;
}

```