#include <bits/stdc++.h>
using namespace std;
const int MX = 200000, INF = 1000000000;
int a[MX], ololo[MX], olololo[MX];
int main() {
  int n, m, last;
  scanf("%d %d %d", &n, &m, &last);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = abs(x - last);
    last = x;
  }
  a[0] = a[n] = INF;
  ololo[0] = 0;
  for (int i = 1; i < n; i++) {
    int j = i - 1;
    while (a[i] > a[j]) j = ololo[j];
    ololo[i] = j;
  }
  olololo[n] = n;
  for (int i = n - 1; i > 0; i--) {
    int j = i + 1;
    while (a[i] >= a[j]) j = olololo[j];
    olololo[i] = j;
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    long long ans = 0;
    for (int i = l; i < r; i++)
      ans += a[i] * 1LL * (min(olololo[i], r) - i) * 1LL *
             (i - max(ololo[i], l - 1));
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 MX = 200000, INF = 1000000000;
int a[MX], ololo[MX], olololo[MX];
int main() {
  int n, m, last;
  scanf("%d %d %d", &n, &m, &last);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = abs(x - last);
    last = x;
  }
  a[0] = a[n] = INF;
  ololo[0] = 0;
  for (int i = 1; i < n; i++) {
    int j = i - 1;
    while (a[i] > a[j]) j = ololo[j];
    ololo[i] = j;
  }
  olololo[n] = n;
  for (int i = n - 1; i > 0; i--) {
    int j = i + 1;
    while (a[i] >= a[j]) j = olololo[j];
    olololo[i] = j;
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    long long ans = 0;
    for (int i = l; i < r; i++)
      ans += a[i] * 1LL * (min(olololo[i], r) - i) * 1LL *
             (i - max(ololo[i], l - 1));
    printf("%I64d\n", ans);
  }
  return 0;
}

```