#include <bits/stdc++.h>
using namespace std;
struct rec {
  int x, w;
  rec() {}
  rec(int a, int b) {
    x = a;
    w = b;
  }
};
deque<rec> que;
int a[100010], b[100010], l[100010], r[100010];
int n, q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) b[i] = abs(a[i + 1] - a[i]);
  while (!que.empty()) que.pop_back();
  for (int i = 1; i < n; ++i) {
    while (!que.empty() && que.back().w < b[i]) que.pop_back();
    if (que.empty())
      l[i] = 1;
    else
      l[i] = que.back().x + 1;
    que.push_back(rec(i, b[i]));
  }
  while (!que.empty()) que.pop_back();
  for (int i = n - 1; i > 0; --i) {
    while (!que.empty() && que.back().w <= b[i]) que.pop_back();
    if (que.empty())
      r[i] = n - 1;
    else
      r[i] = que.back().x - 1;
    que.push_back(rec(i, b[i]));
  }
  long long ans;
  int x, y;
  for (int i = 1; i <= q; ++i) {
    ans = 0;
    scanf("%d%d", &x, &y);
    for (int j = x; j < y; ++j) {
      ans +=
          (long long)b[j] * (j - max(x, l[j]) + 1) * (min(y - 1, r[j]) - j + 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### 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;
struct rec {
  int x, w;
  rec() {}
  rec(int a, int b) {
    x = a;
    w = b;
  }
};
deque<rec> que;
int a[100010], b[100010], l[100010], r[100010];
int n, q;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) b[i] = abs(a[i + 1] - a[i]);
  while (!que.empty()) que.pop_back();
  for (int i = 1; i < n; ++i) {
    while (!que.empty() && que.back().w < b[i]) que.pop_back();
    if (que.empty())
      l[i] = 1;
    else
      l[i] = que.back().x + 1;
    que.push_back(rec(i, b[i]));
  }
  while (!que.empty()) que.pop_back();
  for (int i = n - 1; i > 0; --i) {
    while (!que.empty() && que.back().w <= b[i]) que.pop_back();
    if (que.empty())
      r[i] = n - 1;
    else
      r[i] = que.back().x - 1;
    que.push_back(rec(i, b[i]));
  }
  long long ans;
  int x, y;
  for (int i = 1; i <= q; ++i) {
    ans = 0;
    scanf("%d%d", &x, &y);
    for (int j = x; j < y; ++j) {
      ans +=
          (long long)b[j] * (j - max(x, l[j]) + 1) * (min(y - 1, r[j]) - j + 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```