#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 10;
int n, t;
long long a[maxN];
long long f[maxN], _l[maxN], _r[maxN];
stack<long long> st;
void io() {
  freopen(
      ""
      ".inp",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
int main() {
  cin >> n >> t;
  for (int i = (1), _b = (n); i <= _b; ++i) cin >> a[i];
  for (int i = (1), _b = (n - 1); i <= _b; ++i) f[i] = abs(a[i + 1] - a[i]);
  for (int i = (1), _b = (n - 1); i <= _b; ++i) _l[i] = 1, _r[i] = n - 1;
  f[0] = 1e9;
  st.push(0);
  for (int i = (1), _b = (n - 1); i <= _b; ++i) {
    while (st.size() && f[st.top()] <= f[i]) st.pop();
    if (st.size()) _l[i] = st.top();
    st.push(i);
  }
  while (st.size()) st.pop();
  f[n] = 1e9;
  st.push(n);
  for (int i = (n - 1), _b = (1); i >= _b; --i) {
    while (st.size() && f[st.top()] < f[i]) st.pop();
    if (st.size()) _r[i] = st.top(), st.push(i);
  }
  for (int i = (1), _b = (n - 1); i <= _b; ++i) _l[i]++, _r[i]--;
  int l, r;
  for (int i = (1), _b = (t); i <= _b; ++i) {
    cin >> l >> r;
    long long sum = 0;
    for (int j = (l), _b = (r - 1); j <= _b; ++j) {
      sum += f[j] * (j - max((long long)l, _l[j]) + 1) *
             (min((long long)(r - 1), _r[j]) - j + 1);
    }
    cout << sum << "\n";
  }
  return 0;
}

### 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 + 10;
int n, t;
long long a[maxN];
long long f[maxN], _l[maxN], _r[maxN];
stack<long long> st;
void io() {
  freopen(
      ""
      ".inp",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
int main() {
  cin >> n >> t;
  for (int i = (1), _b = (n); i <= _b; ++i) cin >> a[i];
  for (int i = (1), _b = (n - 1); i <= _b; ++i) f[i] = abs(a[i + 1] - a[i]);
  for (int i = (1), _b = (n - 1); i <= _b; ++i) _l[i] = 1, _r[i] = n - 1;
  f[0] = 1e9;
  st.push(0);
  for (int i = (1), _b = (n - 1); i <= _b; ++i) {
    while (st.size() && f[st.top()] <= f[i]) st.pop();
    if (st.size()) _l[i] = st.top();
    st.push(i);
  }
  while (st.size()) st.pop();
  f[n] = 1e9;
  st.push(n);
  for (int i = (n - 1), _b = (1); i >= _b; --i) {
    while (st.size() && f[st.top()] < f[i]) st.pop();
    if (st.size()) _r[i] = st.top(), st.push(i);
  }
  for (int i = (1), _b = (n - 1); i <= _b; ++i) _l[i]++, _r[i]--;
  int l, r;
  for (int i = (1), _b = (t); i <= _b; ++i) {
    cin >> l >> r;
    long long sum = 0;
    for (int j = (l), _b = (r - 1); j <= _b; ++j) {
      sum += f[j] * (j - max((long long)l, _l[j]) + 1) *
             (min((long long)(r - 1), _r[j]) - j + 1);
    }
    cout << sum << "\n";
  }
  return 0;
}

```