#include <bits/stdc++.h>
using namespace std;
int a[100009];
int d[100009];
int nxt[100009];
int prv[100009];
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) d[i] = abs(a[i] - a[i - 1]);
  vector<int> st;
  st.push_back(1);
  for (int i = 2; i < n; ++i) {
    while (!st.empty() && d[i] > d[st.back()]) {
      nxt[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  while (!st.empty()) {
    nxt[st.back()] = n;
    st.pop_back();
  }
  st.push_back(n - 1);
  for (int i = n - 2; i >= 1; --i) {
    while (!st.empty() && d[i] >= d[st.back()]) {
      prv[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  while (!st.empty()) {
    prv[st.back()] = 0;
    st.pop_back();
  }
  for (int t = 0; t < q; ++t) {
    int l, r;
    cin >> l >> r;
    --r;
    long long ans = 0;
    for (int i = l; i <= r; ++i) {
      int ni = nxt[i];
      int pi = prv[i];
      --ni;
      ++pi;
      if (ni > r) ni = r;
      if (pi < l) pi = l;
      long long cnt = (long long)(i - pi + 1) * (ni - i + 1);
      ans += cnt * d[i];
    }
    cout << ans << '\n';
  }
}

### Prompt

Your task is to create 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;
int a[100009];
int d[100009];
int nxt[100009];
int prv[100009];
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) d[i] = abs(a[i] - a[i - 1]);
  vector<int> st;
  st.push_back(1);
  for (int i = 2; i < n; ++i) {
    while (!st.empty() && d[i] > d[st.back()]) {
      nxt[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  while (!st.empty()) {
    nxt[st.back()] = n;
    st.pop_back();
  }
  st.push_back(n - 1);
  for (int i = n - 2; i >= 1; --i) {
    while (!st.empty() && d[i] >= d[st.back()]) {
      prv[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  while (!st.empty()) {
    prv[st.back()] = 0;
    st.pop_back();
  }
  for (int t = 0; t < q; ++t) {
    int l, r;
    cin >> l >> r;
    --r;
    long long ans = 0;
    for (int i = l; i <= r; ++i) {
      int ni = nxt[i];
      int pi = prv[i];
      --ni;
      ++pi;
      if (ni > r) ni = r;
      if (pi < l) pi = l;
      long long cnt = (long long)(i - pi + 1) * (ni - i + 1);
      ans += cnt * d[i];
    }
    cout << ans << '\n';
  }
}

```