#include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
int n, q;
vector<int> arr, st, diff;
int main() {
  scanf("%d%d", &n, &q);
  arr.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  diff.resize(n - 1);
  for (int i = 0; i < n - 1; i++) {
    diff[i] = abs(arr[i + 1] - arr[i]);
  }
  vector<int> forward(n - 1);
  vector<int> backward(n - 1);
  stack<pair<int, int> > st;
  st.emplace(-1, INF);
  for (int i = 0; i < n - 1; i++) {
    while (diff[i] > st.top().second) st.pop();
    forward[i] = st.top().first;
    st.emplace(i, diff[i]);
  }
  stack<pair<int, int> > st2;
  st2.emplace(n - 1, INF);
  for (int i = n - 2; i >= 0; i--) {
    while (diff[i] >= st2.top().second) st2.pop();
    backward[i] = st2.top().first;
    st2.emplace(i, diff[i]);
  }
  while (q--) {
    int L, R;
    scanf("%d%d", &L, &R);
    L--;
    R--;
    long long res = 0;
    for (int i = L; i < R; i++) {
      int Lptr = max(forward[i], L - 1);
      int Rptr = min(backward[i], R);
      res += (long long)diff[i] * (i - Lptr) * (Rptr - i);
    }
    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;
const int INF = int(1e9);
int n, q;
vector<int> arr, st, diff;
int main() {
  scanf("%d%d", &n, &q);
  arr.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  diff.resize(n - 1);
  for (int i = 0; i < n - 1; i++) {
    diff[i] = abs(arr[i + 1] - arr[i]);
  }
  vector<int> forward(n - 1);
  vector<int> backward(n - 1);
  stack<pair<int, int> > st;
  st.emplace(-1, INF);
  for (int i = 0; i < n - 1; i++) {
    while (diff[i] > st.top().second) st.pop();
    forward[i] = st.top().first;
    st.emplace(i, diff[i]);
  }
  stack<pair<int, int> > st2;
  st2.emplace(n - 1, INF);
  for (int i = n - 2; i >= 0; i--) {
    while (diff[i] >= st2.top().second) st2.pop();
    backward[i] = st2.top().first;
    st2.emplace(i, diff[i]);
  }
  while (q--) {
    int L, R;
    scanf("%d%d", &L, &R);
    L--;
    R--;
    long long res = 0;
    for (int i = L; i < R; i++) {
      int Lptr = max(forward[i], L - 1);
      int Rptr = min(backward[i], R);
      res += (long long)diff[i] * (i - Lptr) * (Rptr - i);
    }
    cout << res << "\n";
  }
  return 0;
}

```