#include <bits/stdc++.h>
using namespace std;
int diff[(100000 + 10)];
pair<int, int> maxi[(100000 + 10)][20];
int LOG;
int h[(100000 + 10)];
int n, q;
int logs[(100000 + 10)];
void preProcess() {
  for (int i = 0; (1 << i) <= n; ++i) logs[1 << i] = i;
  for (int i = 1; i < (100000 + 10); ++i) logs[i] = max(logs[i], logs[i - 1]);
  for (int i = 1; i < n; ++i) {
    diff[i - 1] = abs(h[i] - h[i - 1]);
    maxi[i - 1][0] = {diff[i - 1], i - 1};
  }
  for (LOG = 1; (1 << LOG) <= n; LOG++) {
    for (int i = 0; i + (1 << LOG) <= n - 1; ++i) {
      if (maxi[i][LOG - 1].first >= maxi[i + (1 << (LOG - 1))][LOG - 1].first)
        maxi[i][LOG] = maxi[i][LOG - 1];
      else
        maxi[i][LOG] = maxi[i + (1 << (LOG - 1))][LOG - 1];
    }
  }
}
pair<int, int> get(int l, int len) {
  int p2 = logs[len];
  int r = l + len - 1;
  if (maxi[l][p2].first >= maxi[r - (1 << p2) + 1][p2].first)
    return maxi[l][p2];
  return maxi[r - (1 << p2) + 1][p2];
}
long long solve(int l, int r) {
  if (r == l) return diff[l];
  if (r < l) return 0;
  int len = r - l + 1;
  pair<int, int> pick = get(l, len);
  int idx = pick.second;
  long long ans = solve(l, idx - 1) + solve(idx + 1, r);
  ans += pick.first * 1LL * (idx - l + 1) * (r - idx + 1);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> h[i];
  preProcess();
  while (q--) {
    int l, r;
    cin >> l >> r;
    long long ans = solve(l - 1, r - 2);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 diff[(100000 + 10)];
pair<int, int> maxi[(100000 + 10)][20];
int LOG;
int h[(100000 + 10)];
int n, q;
int logs[(100000 + 10)];
void preProcess() {
  for (int i = 0; (1 << i) <= n; ++i) logs[1 << i] = i;
  for (int i = 1; i < (100000 + 10); ++i) logs[i] = max(logs[i], logs[i - 1]);
  for (int i = 1; i < n; ++i) {
    diff[i - 1] = abs(h[i] - h[i - 1]);
    maxi[i - 1][0] = {diff[i - 1], i - 1};
  }
  for (LOG = 1; (1 << LOG) <= n; LOG++) {
    for (int i = 0; i + (1 << LOG) <= n - 1; ++i) {
      if (maxi[i][LOG - 1].first >= maxi[i + (1 << (LOG - 1))][LOG - 1].first)
        maxi[i][LOG] = maxi[i][LOG - 1];
      else
        maxi[i][LOG] = maxi[i + (1 << (LOG - 1))][LOG - 1];
    }
  }
}
pair<int, int> get(int l, int len) {
  int p2 = logs[len];
  int r = l + len - 1;
  if (maxi[l][p2].first >= maxi[r - (1 << p2) + 1][p2].first)
    return maxi[l][p2];
  return maxi[r - (1 << p2) + 1][p2];
}
long long solve(int l, int r) {
  if (r == l) return diff[l];
  if (r < l) return 0;
  int len = r - l + 1;
  pair<int, int> pick = get(l, len);
  int idx = pick.second;
  long long ans = solve(l, idx - 1) + solve(idx + 1, r);
  ans += pick.first * 1LL * (idx - l + 1) * (r - idx + 1);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> h[i];
  preProcess();
  while (q--) {
    int l, r;
    cin >> l >> r;
    long long ans = solve(l - 1, r - 2);
    cout << ans << endl;
  }
  return 0;
}

```