#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
void solve() {
  int n, q;
  cin >> n >> q;
  vector<int> A(n), B(n), L(n), R(n);
  for (int i = 0; i < (n); ++i) cin >> A[i];
  for (int i = 1; i < n; ++i) B[i] = abs(A[i] - A[i - 1]);
  vector<pair<int, int> > v;
  for (int i = 1; i < n; ++i) {
    int x = B[i];
    while (!v.empty() && x > v.back().first) v.pop_back();
    L[i] = (v.size() > 0 ? v.back().second + 1 : 1);
    v.push_back(make_pair(x, i));
  }
  v.clear();
  for (int i = n - 1; i > 0; --i) {
    int x = B[i];
    while (!v.empty() && x >= v.back().first) v.pop_back();
    R[i] = v.size() > 0 ? v.back().second - 1 : n - 1;
    v.push_back(make_pair(x, i));
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    long long s = 0;
    for (int i = l + 1; i <= r; ++i) {
      int a = i - max(l + 1, L[i]) + 1;
      int b = min(r, R[i]) - i + 1;
      s += 1ll * a * b * B[i];
    }
    cout << s << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Generate 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 N = 1000;
void solve() {
  int n, q;
  cin >> n >> q;
  vector<int> A(n), B(n), L(n), R(n);
  for (int i = 0; i < (n); ++i) cin >> A[i];
  for (int i = 1; i < n; ++i) B[i] = abs(A[i] - A[i - 1]);
  vector<pair<int, int> > v;
  for (int i = 1; i < n; ++i) {
    int x = B[i];
    while (!v.empty() && x > v.back().first) v.pop_back();
    L[i] = (v.size() > 0 ? v.back().second + 1 : 1);
    v.push_back(make_pair(x, i));
  }
  v.clear();
  for (int i = n - 1; i > 0; --i) {
    int x = B[i];
    while (!v.empty() && x >= v.back().first) v.pop_back();
    R[i] = v.size() > 0 ? v.back().second - 1 : n - 1;
    v.push_back(make_pair(x, i));
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    long long s = 0;
    for (int i = l + 1; i <= r; ++i) {
      int a = i - max(l + 1, L[i]) + 1;
      int b = min(r, R[i]) - i + 1;
      s += 1ll * a * b * B[i];
    }
    cout << s << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```