#include <bits/stdc++.h>
int64_t a[100000];
std::pair<int64_t, int> d[100000];
int rb[100000];
int lb[100000];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, q;
  std::cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    if (i > 0) d[i - 1] = std::make_pair(std::abs(a[i] - a[i - 1]), i - 1);
  }
  std::sort(d, d + n - 1);
  std::set<int> b;
  b.insert(-1);
  b.insert(n - 1);
  for (int i = n - 2; i >= 0; --i) {
    int64_t v = d[i].first;
    int64_t k = d[i].second;
    auto it = b.lower_bound(k);
    rb[k] = (*it - 1);
    it--;
    lb[k] = (*it + 1);
    b.insert(k);
  }
  std::sort(d, d + n - 1,
            [](const std::pair<int64_t, int>& lhs,
               const std::pair<int64_t, int>& rhs) {
              return lhs.second < rhs.second;
            });
  for (int qq = 0; qq < q; ++qq) {
    int l, r;
    std::cin >> l >> r;
    l--;
    r--;
    int64_t total = 0;
    for (int i = r - 1; i >= l; --i) {
      auto k = d[i].second;
      total += d[i].first * (k - std::max(lb[k], l) + 1) *
               (std::min(rb[k], r - 1) - k + 1);
    }
    std::cout << total << '\n';
  }
  return 0;
}

### Prompt

Generate 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>
int64_t a[100000];
std::pair<int64_t, int> d[100000];
int rb[100000];
int lb[100000];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, q;
  std::cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    if (i > 0) d[i - 1] = std::make_pair(std::abs(a[i] - a[i - 1]), i - 1);
  }
  std::sort(d, d + n - 1);
  std::set<int> b;
  b.insert(-1);
  b.insert(n - 1);
  for (int i = n - 2; i >= 0; --i) {
    int64_t v = d[i].first;
    int64_t k = d[i].second;
    auto it = b.lower_bound(k);
    rb[k] = (*it - 1);
    it--;
    lb[k] = (*it + 1);
    b.insert(k);
  }
  std::sort(d, d + n - 1,
            [](const std::pair<int64_t, int>& lhs,
               const std::pair<int64_t, int>& rhs) {
              return lhs.second < rhs.second;
            });
  for (int qq = 0; qq < q; ++qq) {
    int l, r;
    std::cin >> l >> r;
    l--;
    r--;
    int64_t total = 0;
    for (int i = r - 1; i >= l; --i) {
      auto k = d[i].second;
      total += d[i].first * (k - std::max(lb[k], l) + 1) *
               (std::min(rb[k], r - 1) - k + 1);
    }
    std::cout << total << '\n';
  }
  return 0;
}

```