#include <bits/stdc++.h>
using namespace std;
const int N = 100001;
int a[N];
int d[N];
int hi[N][20];
int getmax(int s, int t) {
  if (s == t) return s;
  int k = 0;
  while (t - s >= (1 << k)) {
    k++;
  }
  k--;
  int h = s + (1 << k) - 1;
  if (d[hi[t][k]] > d[hi[h][k]]) {
    return hi[t][k];
  } else {
    return hi[h][k];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(15);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    hi[i][0] = i;
  }
  for (int i = 0; i < n - 1; i++) {
    d[i] = abs(a[i + 1] - a[i]);
  }
  for (int k = 1; k < 20; k++) {
    for (int i = 0; i < n - 1; i++) {
      int j = max(i - (1 << (k - 1)), 0);
      if (d[hi[i][k - 1]] < d[hi[j][k - 1]]) {
        hi[i][k] = hi[j][k - 1];
      } else {
        hi[i][k] = hi[i][k - 1];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int s, t;
    cin >> s >> t;
    s--;
    t--;
    queue<pair<int, int> > qu;
    qu.push(make_pair(s, t));
    long long result = 0;
    while (!qu.empty()) {
      pair<int, int> p = qu.front();
      qu.pop();
      s = p.first;
      t = p.second;
      int m = getmax(s, t - 1);
      result += ((long long)(m + 1 - s)) * (long long)(t - m) * d[m];
      if (s != m) qu.push(make_pair(s, m));
      if (m + 1 != t) qu.push(make_pair(m + 1, t));
    }
    cout << result << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 100001;
int a[N];
int d[N];
int hi[N][20];
int getmax(int s, int t) {
  if (s == t) return s;
  int k = 0;
  while (t - s >= (1 << k)) {
    k++;
  }
  k--;
  int h = s + (1 << k) - 1;
  if (d[hi[t][k]] > d[hi[h][k]]) {
    return hi[t][k];
  } else {
    return hi[h][k];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(15);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    hi[i][0] = i;
  }
  for (int i = 0; i < n - 1; i++) {
    d[i] = abs(a[i + 1] - a[i]);
  }
  for (int k = 1; k < 20; k++) {
    for (int i = 0; i < n - 1; i++) {
      int j = max(i - (1 << (k - 1)), 0);
      if (d[hi[i][k - 1]] < d[hi[j][k - 1]]) {
        hi[i][k] = hi[j][k - 1];
      } else {
        hi[i][k] = hi[i][k - 1];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int s, t;
    cin >> s >> t;
    s--;
    t--;
    queue<pair<int, int> > qu;
    qu.push(make_pair(s, t));
    long long result = 0;
    while (!qu.empty()) {
      pair<int, int> p = qu.front();
      qu.pop();
      s = p.first;
      t = p.second;
      int m = getmax(s, t - 1);
      result += ((long long)(m + 1 - s)) * (long long)(t - m) * d[m];
      if (s != m) qu.push(make_pair(s, m));
      if (m + 1 != t) qu.push(make_pair(m + 1, t));
    }
    cout << result << endl;
  }
  return 0;
}

```