#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const long long LINF = (1LL << 62);
const int INF = 1 << 30;
const int NS = 100010;
const int MS = 19;
const int MOD = 1000000007;
const double PI = acos(-1.0);
int n, m, q, h[NS];
pair<int, int> s[NS];
int L[NS], R[NS];
int solve() {}
int main() {
  while (~scanf("%d %d", &n, &q)) {
    m = n - 1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &h[i]);
      if (i > 1) {
        h[i - 1] = abs(h[i - 1] - h[i]);
        s[i - 1] = make_pair(h[i - 1], i - 1);
      }
    }
    sort(s + 1, s + m + 1);
    set<int> idSet;
    idSet.insert(0);
    idSet.insert(n);
    for (int i = m; i >= 1; i--) {
      int pos = s[i].second;
      idSet.insert(pos);
      __typeof(idSet.begin()) it = idSet.find(pos);
      L[pos] = *(--it);
      it++, it++;
      R[pos] = *(it);
    }
    int x, y;
    for (int qi = 0; qi < q; qi++) {
      scanf("%d %d", &x, &y);
      long long ans = 0;
      for (int i = x; i < y; i++) {
        int lb = i - max(L[i], x - 1);
        int rb = min(R[i], y) - i;
        ans += 1LL * h[i] * lb * rb;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const long long LINF = (1LL << 62);
const int INF = 1 << 30;
const int NS = 100010;
const int MS = 19;
const int MOD = 1000000007;
const double PI = acos(-1.0);
int n, m, q, h[NS];
pair<int, int> s[NS];
int L[NS], R[NS];
int solve() {}
int main() {
  while (~scanf("%d %d", &n, &q)) {
    m = n - 1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &h[i]);
      if (i > 1) {
        h[i - 1] = abs(h[i - 1] - h[i]);
        s[i - 1] = make_pair(h[i - 1], i - 1);
      }
    }
    sort(s + 1, s + m + 1);
    set<int> idSet;
    idSet.insert(0);
    idSet.insert(n);
    for (int i = m; i >= 1; i--) {
      int pos = s[i].second;
      idSet.insert(pos);
      __typeof(idSet.begin()) it = idSet.find(pos);
      L[pos] = *(--it);
      it++, it++;
      R[pos] = *(it);
    }
    int x, y;
    for (int qi = 0; qi < q; qi++) {
      scanf("%d %d", &x, &y);
      long long ans = 0;
      for (int i = x; i < y; i++) {
        int lb = i - max(L[i], x - 1);
        int rb = min(R[i], y) - i;
        ans += 1LL * h[i] * lb * rb;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```