B. Split Sort

You are given a permutation$^{\dagger}$ $p_1, p_2, \ldots, p_n$ of integers $1$ to $n$.

You can change the current permutation by applying the following operation several (possibly, zero) times:

• choose some $x$ ($2 \le x \le n$);
• create a new permutation by:
  • first, writing down all elements of $p$ that are less than $x$, without changing their order;
  • second, writing down all elements of $p$ that are greater than or equal to $x$, without changing their order;
• replace $p$ with the newly created permutation.

For example, if the permutation used to be $[6, 4, 3, 5, 2, 1]$ and you choose $x = 4$, then you will first write down $[3, 2, 1]$, then append this with $[6, 4, 5]$. So the initial permutation will be replaced by $[3, 2, 1, 6, 4, 5]$.

Find the minimum number of operations you need to achieve $p_i = i$ for $i = 1, 2, \ldots, n$. We can show that it is always possible to do so.

$^{\dagger}$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($1 \le n \le 100\,000$).

The second line of each test case contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$). It is guaranteed that $p_1, p_2, \ldots, p_n$ is a permutation.

It is guaranteed that the sum of $n$ over all test cases does not exceed $100\,000$.

Output

For each test case, output the answer on a separate line.

Example

input

5
1
1
2
2 1
6
6 4 3 5 2 1
3
3 1 2
19
10 19 7 1 17 11 8 5 12 9 4 18 14 2 6 15 3 16 13

output

0
1
4
1
7

Note

In the first test case, $n = 1$ and $p_1 = 1$, so there is nothing left to do.

In the second test case, we can choose $x = 2$ and we immediately obtain $p_1 = 1$, $p_2 = 2$.

In the third test case, we can achieve the minimum number of operations in the following way:

1. $x = 4$: $[6, 4, 3, 5, 2, 1] \rightarrow [3, 2, 1, 6, 4, 5]$;
2. $x = 6$: $[3, 2, 1, 6, 4, 5] \rightarrow [3, 2, 1, 4, 5, 6]$;
3. $x = 3$: $[3, 2, 1, 4, 5, 6] \rightarrow [2, 1, 3, 4, 5, 6]$;
4. $x = 2$: $[2, 1, 3, 4, 5, 6] \rightarrow [1, 2, 3, 4, 5, 6]$.

解题思路

　　首先很明显$x$可以从$2$枚举到$n$依次进行操作，那么至少$n-1$次就可以将整个序列变成升序。因此很自然会想到某些$x$是否没有必要进行操作。

　　当$x$从$2$枚举到$k$并完成操作后，此时恰好有$a_i = i, \, i \in [1,k-1]$。接着对$x=k+1$进行操作，而改变的地方只有将$a_k$变成$k$，让$k$在$k+1$的前面。如果$k$在$k+1$的前面，是否可以不进行$x=k+1$的操作呢？可以发现如果跳过$x=k+1$而对$x=k+2$进行操作，那么就会得到$a_i = i, \, i \in [1,k+1]$。因此如果发现$k$在$k+1$的前面，那么我们就可以跳过$x=k+1$的操作。而如果$k$在$k+1$的后面，那么必须要进行$x=k+1$的操作，否则永远无法将$k$变到$k+1$的前面。

　　因此记录初始时数组中每个数$i$的下标$p_i$，然后从小到大枚举每个数$i$，如果发现$p_i < p_{i+1}$，说明可以跳过$x=i+1$这个操作，否则进行。

　　AC代码如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 const int N = 1e5 + 10;
 7 
 8 int p[N];
 9 
10 void solve() {
11     int n;
12     scanf("%d", &n);
13     for (int i = 0; i < n; i++) {
14         int x;
15         scanf("%d", &x);
16         p[x] = i;
17     }
18     int ret = 0;
19     for (int i = 1; i < n; i++) {
20         if (p[i + 1] < p[i]) ret++;
21     }
22     printf("%d\n", ret);
23 }
24 
25 int main() {
26     int t;
27     scanf("%d", &t);
28     while (t--) {
29         solve();
30     }
31     
32     return 0;
33 }

参考资料

　　Pinely Round 2 Editorial：https://codeforces.com/blog/entry/119902