#include <bits/stdc++.h>
using namespace std;
int N, Q;
int H[100010] = {0};
int P[100010] = {0};
int pp = 0;
int PL[100010] = {0};
long long delta[100010] = {0};
int main() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    scanf("%d", &H[i]);
    delta[i] = (long long)abs(H[i] - H[i - 1]);
  }
  for (int i = 1; i <= Q; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    pp = 0;
    P[++pp] = L + 1;
    long long Ans = 0;
    for (int j = L + 2; j <= R; j++) {
      for (; pp > 0 && delta[P[pp]] < delta[j]; P[pp--] = 0) {
        if (pp == 1)
          Ans += delta[P[pp]] * (P[pp] - L) * (j - P[pp]);
        else
          Ans += delta[P[pp]] * (P[pp] - P[pp - 1]) * (j - P[pp]);
      }
      P[++pp] = j;
    }
    for (; pp > 0; pp--) {
      if (pp == 1)
        Ans += delta[P[pp]] * (P[pp] - L) * (R - P[pp] + 1);
      else
        Ans += delta[P[pp]] * (P[pp] - P[pp - 1]) * (R - P[pp] + 1);
    }
    cout << Ans << 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, Q;
int H[100010] = {0};
int P[100010] = {0};
int pp = 0;
int PL[100010] = {0};
long long delta[100010] = {0};
int main() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    scanf("%d", &H[i]);
    delta[i] = (long long)abs(H[i] - H[i - 1]);
  }
  for (int i = 1; i <= Q; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    pp = 0;
    P[++pp] = L + 1;
    long long Ans = 0;
    for (int j = L + 2; j <= R; j++) {
      for (; pp > 0 && delta[P[pp]] < delta[j]; P[pp--] = 0) {
        if (pp == 1)
          Ans += delta[P[pp]] * (P[pp] - L) * (j - P[pp]);
        else
          Ans += delta[P[pp]] * (P[pp] - P[pp - 1]) * (j - P[pp]);
      }
      P[++pp] = j;
    }
    for (; pp > 0; pp--) {
      if (pp == 1)
        Ans += delta[P[pp]] * (P[pp] - L) * (R - P[pp] + 1);
      else
        Ans += delta[P[pp]] * (P[pp] - P[pp - 1]) * (R - P[pp] + 1);
    }
    cout << Ans << endl;
  }
  return 0;
}

```