#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
const long long M = 1e9 + 7, inf = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    int len = r - l;
    vector<int> d;
    for (int i = l; i < r; i++) {
      d.push_back(abs(a[i + 1] - a[i]));
    }
    vector<int> ng(len, len), pg(len, -1);
    vector<int> st;
    for (int i = len - 1; i >= 0; i--) {
      while (st.size() and d[i] > d[st.back()]) {
        st.pop_back();
      }
      if (st.size()) {
        ng[i] = st.back();
      }
      st.push_back(i);
    }
    st.clear();
    for (int i = 0; i < len; i++) {
      while (st.size() and d[i] >= d[st.back()]) {
        st.pop_back();
      }
      if (st.size()) {
        pg[i] = st.back();
      }
      st.push_back(i);
    }
    long long ans = 0;
    for (int i = 0; i < len; i++) {
      ans += 1ll * d[i] * (ng[i] - i) * (i - pg[i]);
    }
    cout << ans << "\n";
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
const long long M = 1e9 + 7, inf = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    int len = r - l;
    vector<int> d;
    for (int i = l; i < r; i++) {
      d.push_back(abs(a[i + 1] - a[i]));
    }
    vector<int> ng(len, len), pg(len, -1);
    vector<int> st;
    for (int i = len - 1; i >= 0; i--) {
      while (st.size() and d[i] > d[st.back()]) {
        st.pop_back();
      }
      if (st.size()) {
        ng[i] = st.back();
      }
      st.push_back(i);
    }
    st.clear();
    for (int i = 0; i < len; i++) {
      while (st.size() and d[i] >= d[st.back()]) {
        st.pop_back();
      }
      if (st.size()) {
        pg[i] = st.back();
      }
      st.push_back(i);
    }
    long long ans = 0;
    for (int i = 0; i < len; i++) {
      ans += 1ll * d[i] * (ng[i] - i) * (i - pg[i]);
    }
    cout << ans << "\n";
  }
}

```