#include <bits/stdc++.h>
using namespace std;
int N, Q;
int A[101010];
pair<int, int> P[101010];
int L[101], R[101];
signed long long ret[101];
set<int> S;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N >> Q;
  for (i = 0; i < N; i++) cin >> A[i];
  for (i = 0; i < N - 1; i++) P[i] = {abs(A[i + 1] - A[i]), i};
  sort(P, P + (N - 1));
  reverse(P, P + (N - 1));
  for (i = 0; i < Q; i++) {
    cin >> L[i] >> R[i];
    L[i]--, R[i]--;
  }
  S.insert(-1);
  S.insert(N - 1);
  for (i = 0; i < N - 1; i++) {
    x = P[i].second;
    auto it = S.lower_bound(x);
    int RR = *it;
    it--;
    int LL = 1 + (*it);
    for (j = 0; j < Q; j++) {
      signed long long NR = min(RR, R[j]);
      signed long long NL = max(LL, L[j]);
      if (NL <= x && x + 1 <= NR) {
        signed long long num = (x - NL + 1) * (NR - x);
        ret[j] += num * P[i].first;
      }
    }
    S.insert(x);
  }
  for (j = 0; j < Q; j++) cout << ret[j] << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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[101010];
pair<int, int> P[101010];
int L[101], R[101];
signed long long ret[101];
set<int> S;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N >> Q;
  for (i = 0; i < N; i++) cin >> A[i];
  for (i = 0; i < N - 1; i++) P[i] = {abs(A[i + 1] - A[i]), i};
  sort(P, P + (N - 1));
  reverse(P, P + (N - 1));
  for (i = 0; i < Q; i++) {
    cin >> L[i] >> R[i];
    L[i]--, R[i]--;
  }
  S.insert(-1);
  S.insert(N - 1);
  for (i = 0; i < N - 1; i++) {
    x = P[i].second;
    auto it = S.lower_bound(x);
    int RR = *it;
    it--;
    int LL = 1 + (*it);
    for (j = 0; j < Q; j++) {
      signed long long NR = min(RR, R[j]);
      signed long long NL = max(LL, L[j]);
      if (NL <= x && x + 1 <= NR) {
        signed long long num = (x - NL + 1) * (NR - x);
        ret[j] += num * P[i].first;
      }
    }
    S.insert(x);
  }
  for (j = 0; j < Q; j++) cout << ret[j] << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}

```