#include <bits/stdc++.h>
using namespace std;
using LD = long double;
const int N = 1e5 + 7, inf = 1e9 + 7, MOD = 998244353;
const long long INF = 2e18 + 7;
const int M = 1 << 17;
int dane[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> dane[i];
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    int len = r - l;
    vector<int> roznice;
    roznice.reserve(len);
    for (int i = l; i < r; ++i) {
      roznice.push_back(abs(dane[i + 1] - dane[i]));
    }
    long long res = 0;
    stack<pair<int, int> > kolejka;
    for (int i = 0; i < roznice.size(); ++i) {
      int ileGrup = 1;
      while (kolejka.size() && kolejka.top().first <= roznice[i]) {
        ileGrup += kolejka.top().second;
        res -= (kolejka.top().second) * 1ll * (len - i) * kolejka.top().first;
        kolejka.pop();
      }
      res += roznice[i] * 1ll * ileGrup * (len - i);
      kolejka.push({roznice[i], ileGrup});
    }
    cout << res << '\n';
  }
  return 0;
}

### 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;
using LD = long double;
const int N = 1e5 + 7, inf = 1e9 + 7, MOD = 998244353;
const long long INF = 2e18 + 7;
const int M = 1 << 17;
int dane[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> dane[i];
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    int len = r - l;
    vector<int> roznice;
    roznice.reserve(len);
    for (int i = l; i < r; ++i) {
      roznice.push_back(abs(dane[i + 1] - dane[i]));
    }
    long long res = 0;
    stack<pair<int, int> > kolejka;
    for (int i = 0; i < roznice.size(); ++i) {
      int ileGrup = 1;
      while (kolejka.size() && kolejka.top().first <= roznice[i]) {
        ileGrup += kolejka.top().second;
        res -= (kolejka.top().second) * 1ll * (len - i) * kolejka.top().first;
        kolejka.pop();
      }
      res += roznice[i] * 1ll * ileGrup * (len - i);
      kolejka.push({roznice[i], ileGrup});
    }
    cout << res << '\n';
  }
  return 0;
}

```