#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5, L = 18;
int a[MAXN], N, Q;
pair<int, int> t[MAXN][L];
pair<int, int> query(int i, int j) {
  j++;
  int l = 31 - __builtin_clz(j - i);
  return max(t[i][l], t[j - (1 << l)][l]);
}
long long solve(int i, int j) {
  if (i > j) return 0;
  int k = query(i, j).second;
  return 1ll * (k - i + 1) * (j - k + 1) * a[k] + solve(i, k - 1) +
         solve(k + 1, j);
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 0; i < N; i++) scanf("%d", &a[i]);
  N--;
  for (int i = 0; i < N; i++) {
    a[i] = abs(a[i + 1] - a[i]);
    t[i][0] = {a[i], i};
  }
  for (int l = 1; l < L; l++)
    for (int i = 0; i < N; i++)
      t[i][l] = max(t[i][l - 1], t[i + (1 << (l - 1))][l - 1]);
  for (int i, j; Q--;) {
    scanf("%d%d", &i, &j);
    printf("%I64d\n", solve(i - 1, j - 2));
  }
}

### Prompt

Create a solution in Cpp for 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 MAXN = 2e5, L = 18;
int a[MAXN], N, Q;
pair<int, int> t[MAXN][L];
pair<int, int> query(int i, int j) {
  j++;
  int l = 31 - __builtin_clz(j - i);
  return max(t[i][l], t[j - (1 << l)][l]);
}
long long solve(int i, int j) {
  if (i > j) return 0;
  int k = query(i, j).second;
  return 1ll * (k - i + 1) * (j - k + 1) * a[k] + solve(i, k - 1) +
         solve(k + 1, j);
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 0; i < N; i++) scanf("%d", &a[i]);
  N--;
  for (int i = 0; i < N; i++) {
    a[i] = abs(a[i + 1] - a[i]);
    t[i][0] = {a[i], i};
  }
  for (int l = 1; l < L; l++)
    for (int i = 0; i < N; i++)
      t[i][l] = max(t[i][l - 1], t[i + (1 << (l - 1))][l - 1]);
  for (int i, j; Q--;) {
    scanf("%d%d", &i, &j);
    printf("%I64d\n", solve(i - 1, j - 2));
  }
}

```