#include <bits/stdc++.h>
using namespace std;
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ",";
  _dbg(sdbg + 1, t...);
}
const int N = 1e5 + 5;
int n, q, a[N];
int d[N], L[N], R[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i < n; ++i) {
    d[i] = abs(a[i] - a[i + 1]);
  }
  stack<int> s;
  for (int i = 1; i <= n; ++i) {
    while (s.size() && d[s.top()] <= d[i]) {
      s.pop();
    }
    L[i] = s.empty() ? 0 : s.top();
    s.push(i);
  }
  while (s.size()) {
    s.pop();
  }
  for (int i = n; i >= 1; --i) {
    while (s.size() && d[s.top()] < d[i]) {
      s.pop();
    }
    R[i] = s.empty() ? n : s.top();
    s.push(i);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    int64_t ans = 0;
    for (int i = l; i < r; ++i) {
      int x = max(l, L[i] + 1), y = min(r - 1, R[i] - 1);
      if (x <= y) {
        ans += (int64_t)d[i] * (i - x + 1) * (y - i + 1);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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;
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ",";
  _dbg(sdbg + 1, t...);
}
const int N = 1e5 + 5;
int n, q, a[N];
int d[N], L[N], R[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i < n; ++i) {
    d[i] = abs(a[i] - a[i + 1]);
  }
  stack<int> s;
  for (int i = 1; i <= n; ++i) {
    while (s.size() && d[s.top()] <= d[i]) {
      s.pop();
    }
    L[i] = s.empty() ? 0 : s.top();
    s.push(i);
  }
  while (s.size()) {
    s.pop();
  }
  for (int i = n; i >= 1; --i) {
    while (s.size() && d[s.top()] < d[i]) {
      s.pop();
    }
    R[i] = s.empty() ? n : s.top();
    s.push(i);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    int64_t ans = 0;
    for (int i = l; i < r; ++i) {
      int x = max(l, L[i] + 1), y = min(r - 1, R[i] - 1);
      if (x <= y) {
        ans += (int64_t)d[i] * (i - x + 1) * (y - i + 1);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```