#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n, qn;
  cin >> n >> qn;
  vector<int> a(n);
  vector<array<int, 2>> dif(n - 1);
  for (auto i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (auto i = 0; i < n - 1; ++i) {
    dif[i] = {abs(a[i + 1] - a[i]), i};
  }
  sort(dif.rbegin(), dif.rend());
  set<array<int, 2>> segs{{0, n - 1}};
  vector<array<int, 2>> range(n - 1, {-1, -1});
  for (auto [x, i] : dif) {
    auto it = prev(segs.upper_bound({i, numeric_limits<int>::max()}));
    auto [l, r] = *it;
    segs.erase(it);
    range[i] = {l, r};
    if (l < i) {
      segs.insert({l, i});
    }
    if (i + 1 < r) {
      segs.insert({i + 1, r});
    }
  }
  for (auto qi = 0; qi < qn; ++qi) {
    int ql, qr;
    cin >> ql >> qr, --ql, --qr;
    long long res = 0;
    for (auto [x, i] : dif) {
      auto [l, r] = range[i];
      if (ql <= i && i < qr) {
        res += 1LL * x * (i - max(ql, l) + 1) * (min(qr, r) - i);
      }
    }
    cout << res << "\n";
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n, qn;
  cin >> n >> qn;
  vector<int> a(n);
  vector<array<int, 2>> dif(n - 1);
  for (auto i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (auto i = 0; i < n - 1; ++i) {
    dif[i] = {abs(a[i + 1] - a[i]), i};
  }
  sort(dif.rbegin(), dif.rend());
  set<array<int, 2>> segs{{0, n - 1}};
  vector<array<int, 2>> range(n - 1, {-1, -1});
  for (auto [x, i] : dif) {
    auto it = prev(segs.upper_bound({i, numeric_limits<int>::max()}));
    auto [l, r] = *it;
    segs.erase(it);
    range[i] = {l, r};
    if (l < i) {
      segs.insert({l, i});
    }
    if (i + 1 < r) {
      segs.insert({i + 1, r});
    }
  }
  for (auto qi = 0; qi < qn; ++qi) {
    int ql, qr;
    cin >> ql >> qr, --ql, --qr;
    long long res = 0;
    for (auto [x, i] : dif) {
      auto [l, r] = range[i];
      if (ql <= i && i < qr) {
        res += 1LL * x * (i - max(ql, l) + 1) * (min(qr, r) - i);
      }
    }
    cout << res << "\n";
  }
  return 0;
}

```