#include <bits/stdc++.h>
using namespace std;
int in[100010];
int n, q;
long long int s[100010];
int s1[100010];
int mns(int a, int b) {
  if (a > b)
    return (a - b);
  else
    return (b - a);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int j = 0; j < n; j++) {
    scanf("%d", in + j + 1);
  }
  for (int j = 0; j < q; j++) {
    int l, r;
    long long int ans = 0;
    scanf("%d%d", &l, &r);
    memset(s, 0, sizeof(s));
    memset(s1, 0, sizeof(s1));
    int top = 0;
    ans = mns(in[l], in[l + 1]);
    s[top] = ans;
    s1[top++] = 1;
    long long int last = ans;
    for (int i = l + 1; i < r; i++) {
      long long int t = (long long int)mns(in[i], in[i + 1]);
      int tn = 1;
      while (t >= s[top - 1] && top) {
        int tmp = s1[--top];
        last -= s1[top] * s[top];
        tn += tmp;
      }
      s[top] = t;
      s1[top++] = tn;
      last += t * tn;
      ans += last;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 in[100010];
int n, q;
long long int s[100010];
int s1[100010];
int mns(int a, int b) {
  if (a > b)
    return (a - b);
  else
    return (b - a);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int j = 0; j < n; j++) {
    scanf("%d", in + j + 1);
  }
  for (int j = 0; j < q; j++) {
    int l, r;
    long long int ans = 0;
    scanf("%d%d", &l, &r);
    memset(s, 0, sizeof(s));
    memset(s1, 0, sizeof(s1));
    int top = 0;
    ans = mns(in[l], in[l + 1]);
    s[top] = ans;
    s1[top++] = 1;
    long long int last = ans;
    for (int i = l + 1; i < r; i++) {
      long long int t = (long long int)mns(in[i], in[i + 1]);
      int tn = 1;
      while (t >= s[top - 1] && top) {
        int tmp = s1[--top];
        last -= s1[top] * s[top];
        tn += tmp;
      }
      s[top] = t;
      s1[top++] = tn;
      last += t * tn;
      ans += last;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```