#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos(0.0);
const int maxn = 1e6 + 66;
int aa[maxn], bb[maxn], Left[maxn], Right[maxn];
int main() {
  int n, q;
  while (scanf("%d %d", &n, &q) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
    for (int i = 1; i <= n - 1; i++) bb[i] = abs(aa[i + 1] - aa[i]);
    stack<pair<int, int> > sta;
    sta.push(pair<int, int>(0, INF));
    for (int i = 1; i <= n - 1; i++) {
      while (!sta.empty() && bb[i] > sta.top().second) sta.pop();
      Left[i] = sta.top().first;
      sta.push(pair<int, int>(i, bb[i]));
    }
    while (!sta.empty()) sta.pop();
    sta.push(pair<int, int>(n, INF));
    for (int i = n - 1; i >= 1; i--) {
      while (!sta.empty() && bb[i] >= sta.top().second) sta.pop();
      Right[i] = sta.top().first;
      sta.push(pair<int, int>(i, bb[i]));
    }
    int l, r;
    for (int i = 1; i <= q; i++) {
      scanf("%d%d", &l, &r);
      long long res = 0;
      for (int j = l; j <= r - 1; j++) {
        int a = max(Left[j], l - 1);
        int b = min(Right[j], r);
        res += (long long)bb[j] * (j - a) * (b - j);
      }
      printf("%I64d\n", res);
    }
  }
  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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos(0.0);
const int maxn = 1e6 + 66;
int aa[maxn], bb[maxn], Left[maxn], Right[maxn];
int main() {
  int n, q;
  while (scanf("%d %d", &n, &q) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
    for (int i = 1; i <= n - 1; i++) bb[i] = abs(aa[i + 1] - aa[i]);
    stack<pair<int, int> > sta;
    sta.push(pair<int, int>(0, INF));
    for (int i = 1; i <= n - 1; i++) {
      while (!sta.empty() && bb[i] > sta.top().second) sta.pop();
      Left[i] = sta.top().first;
      sta.push(pair<int, int>(i, bb[i]));
    }
    while (!sta.empty()) sta.pop();
    sta.push(pair<int, int>(n, INF));
    for (int i = n - 1; i >= 1; i--) {
      while (!sta.empty() && bb[i] >= sta.top().second) sta.pop();
      Right[i] = sta.top().first;
      sta.push(pair<int, int>(i, bb[i]));
    }
    int l, r;
    for (int i = 1; i <= q; i++) {
      scanf("%d%d", &l, &r);
      long long res = 0;
      for (int j = l; j <= r - 1; j++) {
        int a = max(Left[j], l - 1);
        int b = min(Right[j], r);
        res += (long long)bb[j] * (j - a) * (b - j);
      }
      printf("%I64d\n", res);
    }
  }
  return 0;
}

```