#include <bits/stdc++.h>
using namespace std;
const int mn = 1e5 + 10;
long long a[mn];
struct comp {
  bool operator()(int l, int r) { return a[l] < a[r]; }
};
int l[mn], r[mn];
long long ans[mn];
int main() {
  int n, q, i;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    scanf("%lld", a + i);
  }
  for (i = 0; i < n - 1; i++) {
    a[i] = abs(a[i] - a[i + 1]);
  }
  for (i = 0; i < q; i++) scanf("%d%d", l + i, r + i), r[i] -= 2, l[i]--;
  priority_queue<int, vector<int>, comp> pq;
  set<int> s;
  s.insert(-1);
  s.insert(n - 1);
  for (i = 0; i < n - 1; i++) pq.push(i);
  while (pq.size()) {
    int x = pq.top();
    pq.pop();
    auto it = s.lower_bound(x);
    int hi = *it, lo = *(--it);
    for (i = 0; i < q; i++) {
      if (x > r[i] || x < l[i]) continue;
      ans[i] += a[x] * (min(hi, r[i] + 1) - x) * (x - max(lo, l[i] - 1));
    }
    s.insert(x);
  }
  for (i = 0; i < q; i++) printf("%lld\n", ans[i]);
}

### 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;
const int mn = 1e5 + 10;
long long a[mn];
struct comp {
  bool operator()(int l, int r) { return a[l] < a[r]; }
};
int l[mn], r[mn];
long long ans[mn];
int main() {
  int n, q, i;
  scanf("%d%d", &n, &q);
  for (i = 0; i < n; i++) {
    scanf("%lld", a + i);
  }
  for (i = 0; i < n - 1; i++) {
    a[i] = abs(a[i] - a[i + 1]);
  }
  for (i = 0; i < q; i++) scanf("%d%d", l + i, r + i), r[i] -= 2, l[i]--;
  priority_queue<int, vector<int>, comp> pq;
  set<int> s;
  s.insert(-1);
  s.insert(n - 1);
  for (i = 0; i < n - 1; i++) pq.push(i);
  while (pq.size()) {
    int x = pq.top();
    pq.pop();
    auto it = s.lower_bound(x);
    int hi = *it, lo = *(--it);
    for (i = 0; i < q; i++) {
      if (x > r[i] || x < l[i]) continue;
      ans[i] += a[x] * (min(hi, r[i] + 1) - x) * (x - max(lo, l[i] - 1));
    }
    s.insert(x);
  }
  for (i = 0; i < q; i++) printf("%lld\n", ans[i]);
}

```