#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v);
void print_array(int* array, int start, int end);
long long a[100100], diff[100100], n, q, lsmall[100100], rsmall[100100], x, y;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = (0), _b = (n); i < _b; i++) cin >> a[i];
  for (int i = (0), _b = (n - 1); i < _b; i++) diff[i] = abs(a[i + 1] - a[i]);
  stack<pair<int, int> > s, s2;
  for (int i = (0), _b = (n - 1); i < _b; i++) {
    while (!s.empty() && diff[i] >= s.top().first) s.pop();
    if (s.empty())
      lsmall[i] = i;
    else
      lsmall[i] = i - 1 - s.top().second;
    s.push(pair<int, int>(diff[i], i));
  }
  for (int i = n - 1; i >= 0; i--) {
    while (!s2.empty() && diff[i] > s2.top().first) s2.pop();
    if (s2.empty())
      rsmall[i] = n - 1 - i;
    else
      rsmall[i] = s2.top().second - i - 1;
    s2.push(pair<int, int>(diff[i], i));
  }
  for (int i = (0), _b = (q); i < _b; i++) {
    long long ans = 0;
    cin >> x >> y;
    x--, y--;
    for (int i = (x), _b = (y); i < _b; i++) {
      long long l = min(lsmall[i], i - x);
      long long r = min(rsmall[i], y - 1 - i);
      ans += (l + 1) * (r + 1) * diff[i];
    }
    cout << ans << endl;
  }
  return 0;
}
void print_array(int* array, int start, int end) {
  printf("[");
  for (int i = (start), _b = (end); i < _b; i++) {
    printf("%d ", array[i]);
  }
  printf("]");
  printf("\n");
}
void print_vector(vector<int> v) {
  printf("[");
  for (int i = (0), _b = (v.size()); i < _b; i++) {
    printf("%d ", v[i]);
  }
  printf("]");
  printf("\n");
}

### Prompt

Please formulate 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;
void print_vector(vector<int> v);
void print_array(int* array, int start, int end);
long long a[100100], diff[100100], n, q, lsmall[100100], rsmall[100100], x, y;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = (0), _b = (n); i < _b; i++) cin >> a[i];
  for (int i = (0), _b = (n - 1); i < _b; i++) diff[i] = abs(a[i + 1] - a[i]);
  stack<pair<int, int> > s, s2;
  for (int i = (0), _b = (n - 1); i < _b; i++) {
    while (!s.empty() && diff[i] >= s.top().first) s.pop();
    if (s.empty())
      lsmall[i] = i;
    else
      lsmall[i] = i - 1 - s.top().second;
    s.push(pair<int, int>(diff[i], i));
  }
  for (int i = n - 1; i >= 0; i--) {
    while (!s2.empty() && diff[i] > s2.top().first) s2.pop();
    if (s2.empty())
      rsmall[i] = n - 1 - i;
    else
      rsmall[i] = s2.top().second - i - 1;
    s2.push(pair<int, int>(diff[i], i));
  }
  for (int i = (0), _b = (q); i < _b; i++) {
    long long ans = 0;
    cin >> x >> y;
    x--, y--;
    for (int i = (x), _b = (y); i < _b; i++) {
      long long l = min(lsmall[i], i - x);
      long long r = min(rsmall[i], y - 1 - i);
      ans += (l + 1) * (r + 1) * diff[i];
    }
    cout << ans << endl;
  }
  return 0;
}
void print_array(int* array, int start, int end) {
  printf("[");
  for (int i = (start), _b = (end); i < _b; i++) {
    printf("%d ", array[i]);
  }
  printf("]");
  printf("\n");
}
void print_vector(vector<int> v) {
  printf("[");
  for (int i = (0), _b = (v.size()); i < _b; i++) {
    printf("%d ", v[i]);
  }
  printf("]");
  printf("\n");
}

```