#include <bits/stdc++.h>
using namespace std;
const int MAX = 100005, INF = 1e9;
int N, Q;
int a[MAX];
int h[MAX];
int l[MAX], r[MAX];
int main(void) {
  cin >> N >> Q;
  for (int i = (1), _b = (N); i <= _b; i++) cin >> a[i];
  for (int i = (2), _b = (N); i <= _b; i++) h[i] = abs(a[i] - a[i - 1]);
  h[1] = h[N + 1] = INF;
  for (int i = (2), _b = (N); i <= _b; i++) {
    l[i] = i;
    while (h[l[i] - 1] <= h[i]) l[i] = l[l[i] - 1];
  }
  for (int i = (N), _b = (2); i >= _b; i--) {
    r[i] = i;
    while (h[r[i] + 1] < h[i]) r[i] = r[r[i] + 1];
  }
  int L, R;
  long long res;
  for (int o = (1), _b = (Q); o <= _b; o++) {
    cin >> L >> R;
    ++L;
    res = 0;
    for (int i = (L), _b = (R); i <= _b; i++) {
      res += (1ll * (min(r[i], R) - i) * (i - max(l[i], L)) + min(r[i], R) -
              max(l[i], L) + 1) *
             h[i];
    }
    cout << res << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAX = 100005, INF = 1e9;
int N, Q;
int a[MAX];
int h[MAX];
int l[MAX], r[MAX];
int main(void) {
  cin >> N >> Q;
  for (int i = (1), _b = (N); i <= _b; i++) cin >> a[i];
  for (int i = (2), _b = (N); i <= _b; i++) h[i] = abs(a[i] - a[i - 1]);
  h[1] = h[N + 1] = INF;
  for (int i = (2), _b = (N); i <= _b; i++) {
    l[i] = i;
    while (h[l[i] - 1] <= h[i]) l[i] = l[l[i] - 1];
  }
  for (int i = (N), _b = (2); i >= _b; i--) {
    r[i] = i;
    while (h[r[i] + 1] < h[i]) r[i] = r[r[i] + 1];
  }
  int L, R;
  long long res;
  for (int o = (1), _b = (Q); o <= _b; o++) {
    cin >> L >> R;
    ++L;
    res = 0;
    for (int i = (L), _b = (R); i <= _b; i++) {
      res += (1ll * (min(r[i], R) - i) * (i - max(l[i], L)) + min(r[i], R) -
              max(l[i], L) + 1) *
             h[i];
    }
    cout << res << "\n";
  }
  return 0;
}

```