#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1e5 + 3;
int n, arr[Maxn], l[Maxn], r[Maxn], num[Maxn];
void findLR(int st, int en);
long long ANS(int st, int en);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 1; i < n; ++i) num[i] = abs(arr[i] - arr[i - 1]);
  while (q--) {
    int st, en;
    cin >> st >> en;
    --st;
    st++;
    cout << ANS(st, en) << endl;
  }
  return 0;
}
void findLR(int st, int en) {
  memset(l, 0, sizeof l);
  memset(r, 0, sizeof r);
  stack<int> q;
  for (int i = st; i < en; ++i) {
    while (!q.empty() && num[i] >= num[q.top()]) {
      int x = q.top();
      q.pop();
      r[x] = i;
    }
    q.push(i);
  }
  while (!q.empty()) {
    r[q.top()] = en;
    q.pop();
  }
  for (int i = en - 1; i >= st; --i) {
    while (!q.empty() && num[i] > num[q.top()]) {
      int x = q.top();
      q.pop();
      l[x] = i + 1;
    }
    q.push(i);
  }
  while (!q.empty()) {
    l[q.top()] = st;
    q.pop();
  }
  return;
}
long long ANS(int st, int en) {
  findLR(st, en);
  long long ans = 0;
  for (int i = st; i < en; ++i)
    ans +=
        (long long)(i - l[i] + 1) * (long long)(r[i] - i) * (long long)num[i];
  return ans;
}

### Prompt

In CPP, your task is to solve 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 Maxn = 1e5 + 3;
int n, arr[Maxn], l[Maxn], r[Maxn], num[Maxn];
void findLR(int st, int en);
long long ANS(int st, int en);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 1; i < n; ++i) num[i] = abs(arr[i] - arr[i - 1]);
  while (q--) {
    int st, en;
    cin >> st >> en;
    --st;
    st++;
    cout << ANS(st, en) << endl;
  }
  return 0;
}
void findLR(int st, int en) {
  memset(l, 0, sizeof l);
  memset(r, 0, sizeof r);
  stack<int> q;
  for (int i = st; i < en; ++i) {
    while (!q.empty() && num[i] >= num[q.top()]) {
      int x = q.top();
      q.pop();
      r[x] = i;
    }
    q.push(i);
  }
  while (!q.empty()) {
    r[q.top()] = en;
    q.pop();
  }
  for (int i = en - 1; i >= st; --i) {
    while (!q.empty() && num[i] > num[q.top()]) {
      int x = q.top();
      q.pop();
      l[x] = i + 1;
    }
    q.push(i);
  }
  while (!q.empty()) {
    l[q.top()] = st;
    q.pop();
  }
  return;
}
long long ANS(int st, int en) {
  findLR(st, en);
  long long ans = 0;
  for (int i = st; i < en; ++i)
    ans +=
        (long long)(i - l[i] + 1) * (long long)(r[i] - i) * (long long)num[i];
  return ans;
}

```