#include <bits/stdc++.h>
using namespace std;
const int N = 110000;
int A[N];
int M[110000][30];
int n, k, x, y, l, r;
void Compute_ST(int N) {
  int i, j;
  for (i = 0; i < N; i++) M[i][0] = i;
  for (j = 1; (1 << j) <= N; j++) {
    for (i = 0; i + (1 << (j - 1)) < N; i++) {
      if (A[M[i][j - 1]] >= A[M[i + (1 << (j - 1))][j - 1]])
        M[i][j] = M[i][j - 1];
      else
        M[i][j] = M[i + (1 << (j - 1))][j - 1];
    }
  }
}
int RMQ(int s, int e) {
  int k = e - s;
  k = 31 - __builtin_clz(k + 1);
  if (A[M[s][k]] >= A[M[e - (1 << k) + 1][k]]) return M[s][k];
  return M[e - (1 << k) + 1][k];
}
long long wynik(int l, int r) {
  int s = RMQ(l, r);
  long long rez = (long long)A[s] * (s - l + 1) * (r - s + 1);
  if (s > l) rez += wynik(l, s - 1);
  if (r > s) rez += wynik(s + 1, r);
  return rez;
}
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n - 1; i++) {
    cin >> y;
    A[i] = abs(x - y);
    x = y;
  }
  Compute_ST(n - 1);
  for (int i = 0; i < k; i++) {
    cin >> l >> r;
    l--;
    r -= 2;
    cout << wynik(l, r) << "\n";
  }
}

### 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 N = 110000;
int A[N];
int M[110000][30];
int n, k, x, y, l, r;
void Compute_ST(int N) {
  int i, j;
  for (i = 0; i < N; i++) M[i][0] = i;
  for (j = 1; (1 << j) <= N; j++) {
    for (i = 0; i + (1 << (j - 1)) < N; i++) {
      if (A[M[i][j - 1]] >= A[M[i + (1 << (j - 1))][j - 1]])
        M[i][j] = M[i][j - 1];
      else
        M[i][j] = M[i + (1 << (j - 1))][j - 1];
    }
  }
}
int RMQ(int s, int e) {
  int k = e - s;
  k = 31 - __builtin_clz(k + 1);
  if (A[M[s][k]] >= A[M[e - (1 << k) + 1][k]]) return M[s][k];
  return M[e - (1 << k) + 1][k];
}
long long wynik(int l, int r) {
  int s = RMQ(l, r);
  long long rez = (long long)A[s] * (s - l + 1) * (r - s + 1);
  if (s > l) rez += wynik(l, s - 1);
  if (r > s) rez += wynik(s + 1, r);
  return rez;
}
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n - 1; i++) {
    cin >> y;
    A[i] = abs(x - y);
    x = y;
  }
  Compute_ST(n - 1);
  for (int i = 0; i < k; i++) {
    cin >> l >> r;
    l--;
    r -= 2;
    cout << wynik(l, r) << "\n";
  }
}

```