Problem:
Score : 2400 points

Problem Statement

Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.

Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.

Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).

How many operations are necessary until the permutation is sorted?

Constraints

1 ≤ N ≤ 2×10^5

(p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print the number of operations that are necessary until the permutation is sorted.

Sample Input 1

5
3 5 1 2 4

Sample Output 1

3

The initial permutation is (3,5,1,2,4), and each operation changes it as follows:

In the first operation, the 1-st and 2-nd elements are high, and the 3-rd, 4-th and 5-th are low. The permutation becomes: (1,2,4,3,5).

In the second operation, the 1-st, 2-nd, 3-rd and 5-th elements are high, and the 4-th is low. The permutation becomes: (3,1,2,4,5).

In the third operation, the 1-st, 4-th and 5-th elements are high, and the 2-nd and 3-rd and 5-th are low. The permutation becomes: (1,2,3,4,5).

Sample Input 2

5
5 4 3 2 1

Sample Output 2

4

Sample Input 3

10
2 10 5 7 3 6 4 9 8 1

Sample Output 3

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <cstdio>
#include <algorithm>
#define maxn 200010
using namespace std;
int n,p[maxn],ans;
int al,bl,a[maxn],b[maxn];
int main() {
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&p[i]);
	ans=0;
	while (1) {
		bool t=true;
		for (int i=1;i<=n;i++)
			if (i!=p[i]) {
				t=false;
				break;
			}
		if (t)
			break;
		ans++;
		int m=0;
		al=0; bl=0;
		for (int i=1;i<=n;i++) {
			m=max(m,p[i]);
			if (m==p[i]) {
				al++;
				a[al]=p[i];
			}
			else {
				bl++;
				b[bl]=p[i];
			}
		}
		for (int i=1;i<=bl;i++)
			p[i]=b[i];
		for (int i=bl+1;i<=n;i++)
			p[i]=a[i-bl];
	}
	printf("%d\n",ans);
	return 0;
}