#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma warning(disable : 4996)
using namespace std;
const long long maxn = 1e5 + 5;
const long long shift = 150;
const long long K = 301;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 1;
long long a[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  while (m--) {
    long long l, r;
    cin >> l >> r;
    vector<long long> b;
    long long ans = 0;
    long long cur = 0;
    for (long long i = l; i < r; i++) b.push_back(abs(a[i] - a[i - 1]));
    vector<pair<long long, long long> > wh;
    for (long long i = 0; i < b.size(); i++) {
      long long nowsz = 1;
      while (!wh.empty() && wh.back().first < b[i]) {
        cur -= wh.back().first * wh.back().second;
        nowsz += wh.back().second;
        wh.pop_back();
      }
      cur += nowsz * b[i];
      ans += cur;
      wh.push_back({b[i], nowsz});
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma warning(disable : 4996)
using namespace std;
const long long maxn = 1e5 + 5;
const long long shift = 150;
const long long K = 301;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 1;
long long a[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  while (m--) {
    long long l, r;
    cin >> l >> r;
    vector<long long> b;
    long long ans = 0;
    long long cur = 0;
    for (long long i = l; i < r; i++) b.push_back(abs(a[i] - a[i - 1]));
    vector<pair<long long, long long> > wh;
    for (long long i = 0; i < b.size(); i++) {
      long long nowsz = 1;
      while (!wh.empty() && wh.back().first < b[i]) {
        cur -= wh.back().first * wh.back().second;
        nowsz += wh.back().second;
        wh.pop_back();
      }
      cur += nowsz * b[i];
      ans += cur;
      wh.push_back({b[i], nowsz});
    }
    cout << ans << '\n';
  }
  return 0;
}

```