#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int n;
int a[100007], d[100007];
int l[100007], r[100007];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) d[i] = abs(a[i + 1] - a[i]);
  stack<pair<int, int> > st;
  for (int i = 1; i < n; i++) {
    while (!st.empty() && st.top().first < d[i]) st.pop();
    if (st.empty())
      l[i] = 0;
    else {
      l[i] = st.top().second;
    }
    st.push(make_pair(d[i], i));
  }
  while (st.empty() == false) st.pop();
  for (int i = n - 1; i >= 1; i--) {
    while (!st.empty() && st.top().first <= d[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else {
      r[i] = st.top().second;
    }
    st.push(make_pair(d[i], i));
  }
  r[n] = n;
  for (int i = 1; i <= q; i++) {
    int lx, rx;
    scanf("%d %d", &lx, &rx);
    long long ans = 0;
    for (int j = lx; j < rx; j++) {
      int wl = max(lx - 1, l[j]);
      int wr = min(rx, r[j]);
      ans += (j - wl) * 1ll * (wr - j) * 1ll * d[j];
    }
    printf("%I64d\n", ans);
  }
  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>
#pragma warning(disable : 4996)
using namespace std;
int n;
int a[100007], d[100007];
int l[100007], r[100007];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) d[i] = abs(a[i + 1] - a[i]);
  stack<pair<int, int> > st;
  for (int i = 1; i < n; i++) {
    while (!st.empty() && st.top().first < d[i]) st.pop();
    if (st.empty())
      l[i] = 0;
    else {
      l[i] = st.top().second;
    }
    st.push(make_pair(d[i], i));
  }
  while (st.empty() == false) st.pop();
  for (int i = n - 1; i >= 1; i--) {
    while (!st.empty() && st.top().first <= d[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else {
      r[i] = st.top().second;
    }
    st.push(make_pair(d[i], i));
  }
  r[n] = n;
  for (int i = 1; i <= q; i++) {
    int lx, rx;
    scanf("%d %d", &lx, &rx);
    long long ans = 0;
    for (int j = lx; j < rx; j++) {
      int wl = max(lx - 1, l[j]);
      int wr = min(rx, r[j]);
      ans += (j - wl) * 1ll * (wr - j) * 1ll * d[j];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```