#include <bits/stdc++.h>
using namespace std;
void solve(istream &in, ostream &out);
int main() {
  ios::sync_with_stdio(false);
  solve(cin, cout);
  return 0;
}
void solve(istream &in, ostream &out) {
  int n, q;
  in >> n >> q;
  int prev;
  in >> prev;
  vector<int> t(n - 1);
  for (int i = 0, next; i < n - 1; ++i) {
    in >> next;
    t[i] = abs(prev - next);
    prev = next;
  }
  vector<int> l(n - 1);
  vector<int> r(n - 1);
  vector<pair<int, int>> s;
  s.emplace_back(numeric_limits<int>::max(), -1);
  for (int i = 0; i < n - 1; ++i) {
    while (s.back() < make_pair(t[i], i)) {
      s.pop_back();
    }
    l[i] = s.back().second + 1;
    s.emplace_back(t[i], i);
  }
  s.clear();
  s.emplace_back(numeric_limits<int>::max(), n - 1);
  for (int i = n - 2; i >= 0; --i) {
    while (s.back() < make_pair(t[i], i)) {
      s.pop_back();
    }
    r[i] = s.back().second - 1;
    s.emplace_back(t[i], i);
  }
  for (int i = 0, a, b; i < q; ++i) {
    in >> a >> b;
    long long result = 0;
    for (long long j = a - 1; j < b - 1; ++j) {
      result += (j - (long long)max(a - 1, l[j]) + 1) *
                ((long long)min(r[j], b - 2) - j + 1) * t[j];
    }
    out << result << endl;
  }
}

### 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;
void solve(istream &in, ostream &out);
int main() {
  ios::sync_with_stdio(false);
  solve(cin, cout);
  return 0;
}
void solve(istream &in, ostream &out) {
  int n, q;
  in >> n >> q;
  int prev;
  in >> prev;
  vector<int> t(n - 1);
  for (int i = 0, next; i < n - 1; ++i) {
    in >> next;
    t[i] = abs(prev - next);
    prev = next;
  }
  vector<int> l(n - 1);
  vector<int> r(n - 1);
  vector<pair<int, int>> s;
  s.emplace_back(numeric_limits<int>::max(), -1);
  for (int i = 0; i < n - 1; ++i) {
    while (s.back() < make_pair(t[i], i)) {
      s.pop_back();
    }
    l[i] = s.back().second + 1;
    s.emplace_back(t[i], i);
  }
  s.clear();
  s.emplace_back(numeric_limits<int>::max(), n - 1);
  for (int i = n - 2; i >= 0; --i) {
    while (s.back() < make_pair(t[i], i)) {
      s.pop_back();
    }
    r[i] = s.back().second - 1;
    s.emplace_back(t[i], i);
  }
  for (int i = 0, a, b; i < q; ++i) {
    in >> a >> b;
    long long result = 0;
    for (long long j = a - 1; j < b - 1; ++j) {
      result += (j - (long long)max(a - 1, l[j]) + 1) *
                ((long long)min(r[j], b - 2) - j + 1) * t[j];
    }
    out << result << endl;
  }
}

```