#include <bits/stdc++.h>
using namespace std;
int a[100006];
int b[100006];
int l[100006];
int r[100006];
long long solve(int n) {
  long long ret = 0;
  int i;
  stack<int> s;
  for (i = 0; i < n; i++) {
    while (!s.empty() && b[s.top()] <= b[i]) s.pop();
    if (s.empty())
      l[i] = 0;
    else
      l[i] = s.top() + 1;
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (i = n - 1; i >= 0; i--) {
    while (!s.empty() && b[s.top()] < b[i]) s.pop();
    if (s.empty())
      r[i] = n;
    else
      r[i] = s.top();
    ret += (long long)(r[i] - i) * (long long)(i - l[i] + 1) * b[i];
    s.push(i);
  }
  return ret;
}
int main() {
  int n, q, i;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--, r--;
    for (i = l; i < r; i++) {
      b[i - l] = abs(a[i] - a[i + 1]);
    }
    long long ans = solve(r - l);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 a[100006];
int b[100006];
int l[100006];
int r[100006];
long long solve(int n) {
  long long ret = 0;
  int i;
  stack<int> s;
  for (i = 0; i < n; i++) {
    while (!s.empty() && b[s.top()] <= b[i]) s.pop();
    if (s.empty())
      l[i] = 0;
    else
      l[i] = s.top() + 1;
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (i = n - 1; i >= 0; i--) {
    while (!s.empty() && b[s.top()] < b[i]) s.pop();
    if (s.empty())
      r[i] = n;
    else
      r[i] = s.top();
    ret += (long long)(r[i] - i) * (long long)(i - l[i] + 1) * b[i];
    s.push(i);
  }
  return ret;
}
int main() {
  int n, q, i;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--, r--;
    for (i = l; i < r; i++) {
      b[i - l] = abs(a[i] - a[i + 1]);
    }
    long long ans = solve(r - l);
    printf("%I64d\n", ans);
  }
  return 0;
}

```