#include <bits/stdc++.h>
using namespace std;
const int M = 100100;
long long mabs(long long x) { return x < 0 ? -x : x; }
int n, m;
long long a[M], h[M];
long long q[M], c[M], qf, qs, sum;
void read() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> h[i];
  }
  for (int i = 0; i + 1 < n; ++i) {
    a[i] = mabs(h[i + 1] - h[i]);
  }
}
void ini() {
  qf = qs = 0;
  sum = 0;
}
void pop() {
  --qf;
  sum -= q[qf] * c[qf];
}
void push(long long x, long long cnt) {
  assert(qs == qf || q[qf - 1] > x);
  sum += x * cnt;
  q[qf] = x;
  c[qf] = cnt;
  ++qf;
}
void add(long long x) {
  long long cur = 1;
  while (qs < qf && q[qf - 1] <= x) {
    pop();
    cur += c[qf];
  }
  push(x, cur);
}
long long get(long long l, long long r) {
  long long ans = 0;
  ini();
  for (int i = l; i < r; ++i) {
    add(a[i]);
    ans += sum;
  }
  return ans;
}
void kill() {
  for (int i = 0; i < m; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << get(l, r) << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.precision(13);
  cout << fixed;
  read();
  kill();
  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;
const int M = 100100;
long long mabs(long long x) { return x < 0 ? -x : x; }
int n, m;
long long a[M], h[M];
long long q[M], c[M], qf, qs, sum;
void read() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> h[i];
  }
  for (int i = 0; i + 1 < n; ++i) {
    a[i] = mabs(h[i + 1] - h[i]);
  }
}
void ini() {
  qf = qs = 0;
  sum = 0;
}
void pop() {
  --qf;
  sum -= q[qf] * c[qf];
}
void push(long long x, long long cnt) {
  assert(qs == qf || q[qf - 1] > x);
  sum += x * cnt;
  q[qf] = x;
  c[qf] = cnt;
  ++qf;
}
void add(long long x) {
  long long cur = 1;
  while (qs < qf && q[qf - 1] <= x) {
    pop();
    cur += c[qf];
  }
  push(x, cur);
}
long long get(long long l, long long r) {
  long long ans = 0;
  ini();
  for (int i = l; i < r; ++i) {
    add(a[i]);
    ans += sum;
  }
  return ans;
}
void kill() {
  for (int i = 0; i < m; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << get(l, r) << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.precision(13);
  cout << fixed;
  read();
  kill();
  return 0;
}

```