#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 6;
const long long INF = 1e15 + 6;
long long a[MAX_N];
long long b[MAX_N];
int main() {
  int n, q;
  while (scanf("%d %d", &n, &q) != EOF) {
    for (int i = 1; n >= i; i++) {
      scanf("%I64d", &b[i]);
      if (i > 1) a[i] = abs(b[i] - b[i - 1]);
    }
    while (q--) {
      int L, R;
      scanf("%d %d", &L, &R);
      L++;
      deque<pair<long long, long long> > dq;
      dq.push_back(make_pair(INF, L - 1));
      long long ans = 0;
      for (long long i = L; R + 1 >= i; i++) {
        long long add = a[i];
        if (i == R + 1) add = INF;
        while (dq.size() > 0 && add > dq[dq.size() - 1].first) {
          ans += ((i - dq[dq.size() - 1].second) *
                  (dq[dq.size() - 1].second - dq[dq.size() - 2].second)) *
                 (dq[dq.size() - 1].first);
          dq.pop_back();
        }
        dq.push_back(make_pair(add, i));
      }
      printf("%I64d\n", ans);
    }
  }
}

### Prompt

In cpp, your task is to solve 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 MAX_N = 1e5 + 6;
const long long INF = 1e15 + 6;
long long a[MAX_N];
long long b[MAX_N];
int main() {
  int n, q;
  while (scanf("%d %d", &n, &q) != EOF) {
    for (int i = 1; n >= i; i++) {
      scanf("%I64d", &b[i]);
      if (i > 1) a[i] = abs(b[i] - b[i - 1]);
    }
    while (q--) {
      int L, R;
      scanf("%d %d", &L, &R);
      L++;
      deque<pair<long long, long long> > dq;
      dq.push_back(make_pair(INF, L - 1));
      long long ans = 0;
      for (long long i = L; R + 1 >= i; i++) {
        long long add = a[i];
        if (i == R + 1) add = INF;
        while (dq.size() > 0 && add > dq[dq.size() - 1].first) {
          ans += ((i - dq[dq.size() - 1].second) *
                  (dq[dq.size() - 1].second - dq[dq.size() - 2].second)) *
                 (dq[dq.size() - 1].first);
          dq.pop_back();
        }
        dq.push_back(make_pair(add, i));
      }
      printf("%I64d\n", ans);
    }
  }
}

```