#include <bits/stdc++.h>
using namespace std;
int N, Q;
int A[100000];
pair<int, int> qs[100];
long long ans[100];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> Q;
  for (int i = 0; i < (N); i++) cin >> A[i];
  for (int i = 0; i < (N - 1); i++) A[i] = abs(A[i + 1] - A[i]);
  A[N - 1] = 0;
  N--;
  vector<pair<int, int> > xs;
  for (int i = 0; i < (N); i++) xs.push_back(pair<int, int>(A[i], i));
  sort((xs).begin(), (xs).end(), greater<pair<int, int> >());
  for (int i = 0; i < (Q); i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    r--;
    qs[i] = pair<int, int>(l, r);
  }
  set<int> g;
  g.insert(-1);
  g.insert(N);
  long long s = 0;
  for (pair<int, int> p : xs) {
    int val = p.first, pos = p.second;
    int left = pos - *(--g.upper_bound(pos));
    int right = *g.lower_bound(pos) - pos;
    g.insert(pos);
    for (int i = 0; i < (Q); i++) {
      int l = qs[i].first, r = qs[i].second;
      if (pos < l || r < pos) continue;
      ans[i] += 1LL * val * min(pos - l + 1, left) * min(right, r - pos + 1);
    }
  }
  for (int i = 0; i < (Q); i++) cout << ans[i] << "\n";
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int N, Q;
int A[100000];
pair<int, int> qs[100];
long long ans[100];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> Q;
  for (int i = 0; i < (N); i++) cin >> A[i];
  for (int i = 0; i < (N - 1); i++) A[i] = abs(A[i + 1] - A[i]);
  A[N - 1] = 0;
  N--;
  vector<pair<int, int> > xs;
  for (int i = 0; i < (N); i++) xs.push_back(pair<int, int>(A[i], i));
  sort((xs).begin(), (xs).end(), greater<pair<int, int> >());
  for (int i = 0; i < (Q); i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    r--;
    qs[i] = pair<int, int>(l, r);
  }
  set<int> g;
  g.insert(-1);
  g.insert(N);
  long long s = 0;
  for (pair<int, int> p : xs) {
    int val = p.first, pos = p.second;
    int left = pos - *(--g.upper_bound(pos));
    int right = *g.lower_bound(pos) - pos;
    g.insert(pos);
    for (int i = 0; i < (Q); i++) {
      int l = qs[i].first, r = qs[i].second;
      if (pos < l || r < pos) continue;
      ans[i] += 1LL * val * min(pos - l + 1, left) * min(right, r - pos + 1);
    }
  }
  for (int i = 0; i < (Q); i++) cout << ans[i] << "\n";
  return 0;
}

```