#include <bits/stdc++.h>
using namespace std;
const int MAX = 105000;
int data[MAX];
long long maxx[MAX];
int pre[MAX];
int nex[MAX];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &data[i]);
    if (i >= 2) {
      maxx[i - 1] = abs(data[i] - data[i - 1]);
    }
  }
  stack<int> stk;
  for (int i = 1; i < n; i++) {
    while (!stk.empty() && maxx[i] >= maxx[stk.top()]) {
      int now = stk.top();
      nex[now] = i;
      stk.pop();
    }
    if (stk.empty())
      pre[i] = 1;
    else {
      pre[i] = stk.top() + 1;
    }
    stk.push(i);
  }
  while (!stk.empty()) {
    int now = stk.top();
    nex[now] = n;
    stk.pop();
  }
  int ta, tb;
  long long ans = 0;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &ta, &tb);
    ans = 0;
    for (int i = ta; i < tb; i++) {
      ans += (long long)(i - max(ta, pre[i]) + 1) * (min(tb, nex[i]) - i) *
             maxx[i];
    }
    cout << ans << endl;
  }
}

### Prompt

Generate 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;
const int MAX = 105000;
int data[MAX];
long long maxx[MAX];
int pre[MAX];
int nex[MAX];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &data[i]);
    if (i >= 2) {
      maxx[i - 1] = abs(data[i] - data[i - 1]);
    }
  }
  stack<int> stk;
  for (int i = 1; i < n; i++) {
    while (!stk.empty() && maxx[i] >= maxx[stk.top()]) {
      int now = stk.top();
      nex[now] = i;
      stk.pop();
    }
    if (stk.empty())
      pre[i] = 1;
    else {
      pre[i] = stk.top() + 1;
    }
    stk.push(i);
  }
  while (!stk.empty()) {
    int now = stk.top();
    nex[now] = n;
    stk.pop();
  }
  int ta, tb;
  long long ans = 0;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &ta, &tb);
    ans = 0;
    for (int i = ta; i < tb; i++) {
      ans += (long long)(i - max(ta, pre[i]) + 1) * (min(tb, nex[i]) - i) *
             maxx[i];
    }
    cout << ans << endl;
  }
}

```