#include <bits/stdc++.h>
using namespace std;
int const LIM = 1e5 + 100;
int n, m, a[LIM], nxt[LIM], prv[LIM];
long long x[LIM];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  stack<int, vector<int>> s;
  for (int i = 0; i < m; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    int count = 0;
    for (int i = l; i + 1 < r; ++i) x[count++] = abs(a[i] - a[i + 1]);
    for (int i = 0; i < count; ++i) {
      while (!s.empty() && x[i] > x[s.top()]) {
        nxt[s.top()] = i;
        s.pop();
      }
      s.push(i);
    }
    while (!s.empty()) {
      nxt[s.top()] = count;
      s.pop();
    }
    for (int i = count - 1; i >= 0; --i) {
      while (!s.empty() && x[i] >= x[s.top()]) {
        prv[s.top()] = i;
        s.pop();
      }
      s.push(i);
    }
    while (!s.empty()) {
      prv[s.top()] = -1;
      s.pop();
    }
    long long answer = 0;
    for (int i = 0; i < count; ++i) {
      answer += x[i] * (1LL * i - prv[i]) * (1LL * nxt[i] - i);
    }
    cout << answer << '\n';
  }
  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;
int const LIM = 1e5 + 100;
int n, m, a[LIM], nxt[LIM], prv[LIM];
long long x[LIM];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  stack<int, vector<int>> s;
  for (int i = 0; i < m; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    int count = 0;
    for (int i = l; i + 1 < r; ++i) x[count++] = abs(a[i] - a[i + 1]);
    for (int i = 0; i < count; ++i) {
      while (!s.empty() && x[i] > x[s.top()]) {
        nxt[s.top()] = i;
        s.pop();
      }
      s.push(i);
    }
    while (!s.empty()) {
      nxt[s.top()] = count;
      s.pop();
    }
    for (int i = count - 1; i >= 0; --i) {
      while (!s.empty() && x[i] >= x[s.top()]) {
        prv[s.top()] = i;
        s.pop();
      }
      s.push(i);
    }
    while (!s.empty()) {
      prv[s.top()] = -1;
      s.pop();
    }
    long long answer = 0;
    for (int i = 0; i < count; ++i) {
      answer += x[i] * (1LL * i - prv[i]) * (1LL * nxt[i] - i);
    }
    cout << answer << '\n';
  }
  return 0;
}

```