#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int arr[N];
const int ST_N = 200000;
int Bin[20], Log[ST_N];
int mn[20][ST_N];
void ST(int n) {
  Bin[0] = 1;
  for (int i = 1; i < 20; i++) Bin[i] = Bin[i - 1] << 1;
  for (int i = 1; i <= n; i++) mn[0][i] = arr[i];
  int k = int(log2(n));
  for (int i = 1; i <= k; i++)
    for (int j = 1; j + Bin[i] - 1 <= n; j++)
      mn[i][j] = max(mn[i - 1][j], mn[i - 1][j + Bin[i - 1]]);
}
int right_max[N], left_max[N];
void init_LR(int n) {
  int k = int(log2(n));
  int temp;
  for (int i = 1; i < n - 1; i++) {
    temp = i + 1;
    for (int j = k; j >= 0; j--)
      if (temp + (1 << j) - 1 < n && mn[j][temp] <= arr[i])
        right_max[i] += 1 << j, temp += 1 << j;
  }
  for (int i = 2; i < n; i++) {
    temp = i - 1;
    for (int j = k; j >= 0; j--)
      if (temp - (1 << j) + 1 >= 1 && mn[j][temp - (1 << j) + 1] < arr[i])
        left_max[i] += 1 << j, temp -= 1 << j;
  }
}
int seq[N];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &seq[i]);
  for (int i = 1; i < n; i++) arr[i] = abs(seq[i + 1] - seq[i]);
  ST(n);
  init_LR(n);
  long long ans = 0;
  int l, r;
  while (q--) {
    scanf("%d %d", &l, &r);
    ans = 0;
    for (int i = l; i < r; i++)
      ans += 1LL * (min(left_max[i], i - l) + 1) *
             (min(right_max[i], r - i - 1) + 1) * arr[i];
    printf("%lld\n", ans);
  }
  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;
const int N = 1e5 + 5;
int arr[N];
const int ST_N = 200000;
int Bin[20], Log[ST_N];
int mn[20][ST_N];
void ST(int n) {
  Bin[0] = 1;
  for (int i = 1; i < 20; i++) Bin[i] = Bin[i - 1] << 1;
  for (int i = 1; i <= n; i++) mn[0][i] = arr[i];
  int k = int(log2(n));
  for (int i = 1; i <= k; i++)
    for (int j = 1; j + Bin[i] - 1 <= n; j++)
      mn[i][j] = max(mn[i - 1][j], mn[i - 1][j + Bin[i - 1]]);
}
int right_max[N], left_max[N];
void init_LR(int n) {
  int k = int(log2(n));
  int temp;
  for (int i = 1; i < n - 1; i++) {
    temp = i + 1;
    for (int j = k; j >= 0; j--)
      if (temp + (1 << j) - 1 < n && mn[j][temp] <= arr[i])
        right_max[i] += 1 << j, temp += 1 << j;
  }
  for (int i = 2; i < n; i++) {
    temp = i - 1;
    for (int j = k; j >= 0; j--)
      if (temp - (1 << j) + 1 >= 1 && mn[j][temp - (1 << j) + 1] < arr[i])
        left_max[i] += 1 << j, temp -= 1 << j;
  }
}
int seq[N];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &seq[i]);
  for (int i = 1; i < n; i++) arr[i] = abs(seq[i + 1] - seq[i]);
  ST(n);
  init_LR(n);
  long long ans = 0;
  int l, r;
  while (q--) {
    scanf("%d %d", &l, &r);
    ans = 0;
    for (int i = l; i < r; i++)
      ans += 1LL * (min(left_max[i], i - l) + 1) *
             (min(right_max[i], r - i - 1) + 1) * arr[i];
    printf("%lld\n", ans);
  }
  return 0;
}

```