#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = ~0U >> 1;
int n, q, a[N], b[N], l[N], r[N];
stack<pair<int, int> > st;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i) {
      b[i - 1] = abs(a[i] - a[i - 1]);
    }
  }
  n--;
  int now = 0;
  st.emplace(-1, inf);
  for (int i = 0; i < (n); i++) {
    while (st.top().second < b[i]) st.pop();
    l[i] = st.top().first + 1;
    st.emplace(i, b[i]);
  }
  st = stack<pair<int, int> >();
  st.emplace(n, inf);
  for (int i = (n - 1); i > (-1); i--) {
    while (st.top().second <= b[i]) st.pop();
    r[i] = st.top().first - 1;
    st.emplace(i, b[i]);
  }
  while (q--) {
    int u, v;
    long long ans = 0;
    cin >> u >> v;
    u--;
    v--;
    for (int i = (u); i < (v); i++) {
      ans += b[i] * 1ll * (i - max(l[i], u) + 1) * (min(r[i], v - 1) - i + 1);
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const int N = 100010;
const int inf = ~0U >> 1;
int n, q, a[N], b[N], l[N], r[N];
stack<pair<int, int> > st;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i) {
      b[i - 1] = abs(a[i] - a[i - 1]);
    }
  }
  n--;
  int now = 0;
  st.emplace(-1, inf);
  for (int i = 0; i < (n); i++) {
    while (st.top().second < b[i]) st.pop();
    l[i] = st.top().first + 1;
    st.emplace(i, b[i]);
  }
  st = stack<pair<int, int> >();
  st.emplace(n, inf);
  for (int i = (n - 1); i > (-1); i--) {
    while (st.top().second <= b[i]) st.pop();
    r[i] = st.top().first - 1;
    st.emplace(i, b[i]);
  }
  while (q--) {
    int u, v;
    long long ans = 0;
    cin >> u >> v;
    u--;
    v--;
    for (int i = (u); i < (v); i++) {
      ans += b[i] * 1ll * (i - max(l[i], u) + 1) * (min(r[i], v - 1) - i + 1);
    }
    cout << ans << endl;
  }
  return 0;
}

```