#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int d[N], dp[N][20];
void pre(int n) {
  for (int i = 1; i <= n; i++) dp[i][0] = i;
  for (int i = 1; ((1 << i) - 1) < n; i++) {
    for (int j = 1; (j + (1 << i) - 1) <= n; j++) {
      dp[j][i] = dp[j][i - 1];
      if (d[dp[j][i - 1]] < d[dp[j + (1 << (i - 1))][i - 1]])
        dp[j][i] = dp[j + (1 << (i - 1))][i - 1];
    }
  }
}
int query(int l, int r) {
  int x = log2(r - l + 1);
  if (d[dp[l][x]] > d[dp[r - (1 << x) + 1][x]]) return dp[l][x];
  return dp[r - (1 << x) + 1][x];
}
long long solve(int l, int r) {
  if (l >= r) return 0;
  int ind = query(l, r - 1);
  long long ans = 1LL * (ind - l + 1) * (r - ind) * d[ind] + solve(l, ind) +
                  solve(ind + 1, r);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q, l, r, n;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> d[i];
  for (int i = 2; i <= n; i++) d[i - 1] = abs(d[i] - d[i - 1]);
  pre(n - 1);
  while (q--) {
    cin >> l >> r;
    cout << solve(l, r) << "\n";
  }
  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 = 100005;
int d[N], dp[N][20];
void pre(int n) {
  for (int i = 1; i <= n; i++) dp[i][0] = i;
  for (int i = 1; ((1 << i) - 1) < n; i++) {
    for (int j = 1; (j + (1 << i) - 1) <= n; j++) {
      dp[j][i] = dp[j][i - 1];
      if (d[dp[j][i - 1]] < d[dp[j + (1 << (i - 1))][i - 1]])
        dp[j][i] = dp[j + (1 << (i - 1))][i - 1];
    }
  }
}
int query(int l, int r) {
  int x = log2(r - l + 1);
  if (d[dp[l][x]] > d[dp[r - (1 << x) + 1][x]]) return dp[l][x];
  return dp[r - (1 << x) + 1][x];
}
long long solve(int l, int r) {
  if (l >= r) return 0;
  int ind = query(l, r - 1);
  long long ans = 1LL * (ind - l + 1) * (r - ind) * d[ind] + solve(l, ind) +
                  solve(ind + 1, r);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q, l, r, n;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> d[i];
  for (int i = 2; i <= n; i++) d[i - 1] = abs(d[i] - d[i - 1]);
  pre(n - 1);
  while (q--) {
    cin >> l >> r;
    cout << solve(l, r) << "\n";
  }
  return 0;
}

```