#include <bits/stdc++.h>
using namespace std;
int A[100010];
struct xx {
  int st, op;
  xx(int a, int b) { st = a, op = b; }
};
deque<xx> Q;
int main() {
  int n, q, bf, now;
  scanf("%d%d", &n, &q);
  scanf("%d", &bf);
  for (int c = 2; c <= n; c++) {
    scanf("%d", &now);
    A[c - 1] = abs(now - bf);
    bf = now;
  }
  for (int cc = 1; cc <= q; cc++) {
    int a, b;
    scanf("%d%d", &a, &b);
    Q.push_back(xx(a, 1));
    long long ans = 0;
    for (int c = a + 1; c <= b - 1; c++) {
      int op = 1;
      while (!Q.empty() and A[Q.back().st] < A[c]) {
        op += Q.back().op;
        ans += (long long)A[Q.back().st] * (c - Q.back().st) * Q.back().op;
        Q.pop_back();
      }
      Q.push_back(xx(c, op));
    }
    while (!Q.empty()) {
      ans += (long long)A[Q.back().st] * (b - Q.back().st) * Q.back().op;
      Q.pop_back();
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Please formulate a cpp solution 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;
int A[100010];
struct xx {
  int st, op;
  xx(int a, int b) { st = a, op = b; }
};
deque<xx> Q;
int main() {
  int n, q, bf, now;
  scanf("%d%d", &n, &q);
  scanf("%d", &bf);
  for (int c = 2; c <= n; c++) {
    scanf("%d", &now);
    A[c - 1] = abs(now - bf);
    bf = now;
  }
  for (int cc = 1; cc <= q; cc++) {
    int a, b;
    scanf("%d%d", &a, &b);
    Q.push_back(xx(a, 1));
    long long ans = 0;
    for (int c = a + 1; c <= b - 1; c++) {
      int op = 1;
      while (!Q.empty() and A[Q.back().st] < A[c]) {
        op += Q.back().op;
        ans += (long long)A[Q.back().st] * (c - Q.back().st) * Q.back().op;
        Q.pop_back();
      }
      Q.push_back(xx(c, op));
    }
    while (!Q.empty()) {
      ans += (long long)A[Q.back().st] * (b - Q.back().st) * Q.back().op;
      Q.pop_back();
    }
    printf("%lld\n", ans);
  }
}

```