#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[211111];
pair<int, int> d[22][211111];
int gg[211111];
long long sol(int x, int y) {
  if (y - x <= 0) return 0;
  int w = gg[y - x];
  int z = max(d[w][x], d[w][(y - 1) - (1ll << (w)) + 1]).second;
  long long ret = abs(a[z] - a[z + 1]) * 1ll * (z - x + 1) * (y - z);
  ret += sol(x, z);
  ret += sol(z + 1, y);
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) d[0][i] = make_pair(abs(a[i] - a[i + 1]), i);
  gg[1] = 0;
  for (int i = 2; i <= n; i++) gg[i] = gg[i / 2] + 1;
  for (int i = 1; i <= gg[n]; i++)
    for (int j = 0; j + (1ll << (i)) <= n - 1; j++)
      d[i][j] = max(d[i - 1][j], d[i - 1][j + (1ll << (i - 1))]);
  for (int i = 0; i < k; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    cout << sol(x, y) << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int n, k;
int a[211111];
pair<int, int> d[22][211111];
int gg[211111];
long long sol(int x, int y) {
  if (y - x <= 0) return 0;
  int w = gg[y - x];
  int z = max(d[w][x], d[w][(y - 1) - (1ll << (w)) + 1]).second;
  long long ret = abs(a[z] - a[z + 1]) * 1ll * (z - x + 1) * (y - z);
  ret += sol(x, z);
  ret += sol(z + 1, y);
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) d[0][i] = make_pair(abs(a[i] - a[i + 1]), i);
  gg[1] = 0;
  for (int i = 2; i <= n; i++) gg[i] = gg[i / 2] + 1;
  for (int i = 1; i <= gg[n]; i++)
    for (int j = 0; j + (1ll << (i)) <= n - 1; j++)
      d[i][j] = max(d[i - 1][j], d[i - 1][j + (1ll << (i - 1))]);
  for (int i = 0; i < k; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    cout << sol(x, y) << endl;
  }
  return 0;
}

```