#include <bits/stdc++.h>
using namespace std;
template <typename input_type>
input_type get() {
  input_type input;
  cin >> input;
  return input;
}
int main() {
  int size = get<int>(), query = get<int>();
  long long int lft[size + 5], rght[size + 5], array[size + 5];
  for (int i = 1; i <= size; i++) {
    array[i] = get<int>();
  }
  for (int i = size; i >= 1; i--) {
    array[i] = abs(array[i] - array[i - 1]);
  }
  lft[2] = 1;
  for (int i = 3; i <= size; i++) {
    int tmp = (i - 1);
    while ((array[tmp] <= array[i]) and tmp) {
      tmp = lft[tmp];
    }
    lft[i] = tmp;
  }
  rght[size] = (size + 1);
  for (int i = (size - 1); i > 1; i--) {
    int j = (i + 1);
    while ((array[j] < array[i]) and (j <= size)) {
      j = rght[j];
    }
    rght[i] = j;
  }
  for (int i = 1; i <= query; i++) {
    long long int ans = 0;
    pair<int, int> endpnt = {(get<int>() + 1), get<int>()};
    for (int j = endpnt.first; j <= endpnt.second; j++) {
      ans += (min((long long int)endpnt.second + 1, rght[j]) - j) *
             (j - max((long long int)endpnt.first - 1, lft[j])) * array[j];
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <typename input_type>
input_type get() {
  input_type input;
  cin >> input;
  return input;
}
int main() {
  int size = get<int>(), query = get<int>();
  long long int lft[size + 5], rght[size + 5], array[size + 5];
  for (int i = 1; i <= size; i++) {
    array[i] = get<int>();
  }
  for (int i = size; i >= 1; i--) {
    array[i] = abs(array[i] - array[i - 1]);
  }
  lft[2] = 1;
  for (int i = 3; i <= size; i++) {
    int tmp = (i - 1);
    while ((array[tmp] <= array[i]) and tmp) {
      tmp = lft[tmp];
    }
    lft[i] = tmp;
  }
  rght[size] = (size + 1);
  for (int i = (size - 1); i > 1; i--) {
    int j = (i + 1);
    while ((array[j] < array[i]) and (j <= size)) {
      j = rght[j];
    }
    rght[i] = j;
  }
  for (int i = 1; i <= query; i++) {
    long long int ans = 0;
    pair<int, int> endpnt = {(get<int>() + 1), get<int>()};
    for (int j = endpnt.first; j <= endpnt.second; j++) {
      ans += (min((long long int)endpnt.second + 1, rght[j]) - j) *
             (j - max((long long int)endpnt.first - 1, lft[j])) * array[j];
    }
    cout << ans << '\n';
  }
  return 0;
}

```