#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
int main() {
  int n, m;
  int i, j, k;
  long long int a, b, c, d;
  vector<long long int> v1, v2, vl, vr;
  priority_queue<pair<int, int> > q1;
  cin >> n >> m;
  cin >> b;
  for (i = 0; i < n - 1; i++) {
    cin >> a;
    q1.push(make_pair(abs(a - b), i));
    b = a;
  }
  n--;
  set<int> s1;
  s1.insert(-1), s1.insert(n);
  for (i = 0; i < n; i++) {
    a = q1.top().first, b = q1.top().second;
    q1.pop();
    v1.push_back(a);
    v2.push_back(b);
    auto itr = s1.lower_bound(b);
    vr.push_back(*itr);
    itr--;
    vl.push_back(*itr);
    s1.insert(b);
  }
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    a -= 2, b--;
    long long int s = 0;
    for (j = 0; j < n; j++) {
      c = max(vl[j], a), d = min(vr[j], b);
      if (c >= v2[j] || v2[j] >= d) continue;
      s += v1[j] * (v2[j] - c) * (d - v2[j]);
    }
    cout << s << endl;
  }
}

### 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 INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
int main() {
  int n, m;
  int i, j, k;
  long long int a, b, c, d;
  vector<long long int> v1, v2, vl, vr;
  priority_queue<pair<int, int> > q1;
  cin >> n >> m;
  cin >> b;
  for (i = 0; i < n - 1; i++) {
    cin >> a;
    q1.push(make_pair(abs(a - b), i));
    b = a;
  }
  n--;
  set<int> s1;
  s1.insert(-1), s1.insert(n);
  for (i = 0; i < n; i++) {
    a = q1.top().first, b = q1.top().second;
    q1.pop();
    v1.push_back(a);
    v2.push_back(b);
    auto itr = s1.lower_bound(b);
    vr.push_back(*itr);
    itr--;
    vl.push_back(*itr);
    s1.insert(b);
  }
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    a -= 2, b--;
    long long int s = 0;
    for (j = 0; j < n; j++) {
      c = max(vl[j], a), d = min(vr[j], b);
      if (c >= v2[j] || v2[j] >= d) continue;
      s += v1[j] * (v2[j] - c) * (d - v2[j]);
    }
    cout << s << endl;
  }
}

```