Ivan wants to play a game with you. He picked some string ssnn

You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 11n−1n−1

Ivan wants you to guess which of the given 2n−22n−2

Input

The first line of the input contains one integer number nn2≤n≤1002≤n≤100ss

The next 2n−22n−211n−1n−1

It is guaranteed that there are exactly 2211n−1n−1nn

Output

Print one string of length 2n−22n−2iiii

If there are several possible answers, you can print any.

Examples

input

Copy

5
ba
a
abab
a
aba
baba
ab
aba

output

Copy

SPPSPSPS

input

Copy

3
a
aa
aa
a

output

Copy

PPSS

input

Copy

2
a
c

output

Copy

PS

Note

The only string which Ivan can guess in the first example is "ababa".

The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.

In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.

思路：字符串暴力匹配，首先将字符串按长度排序，然后任意指定一个最小的为前缀，一个最小的为后缀，从小到大进行匹配，如果匹配不上，就反过来重新指定前缀后缀。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1000;
int n;
string s[maxn];
struct Node {
	string t;
	int num;
}p[maxn];
char c[maxn];
bool cmp(Node x, Node y) {
	return x.t.size() < y.t.size();
}
bool is_last(string tem, string tt) {//后缀匹配
	int len = tem.size();
	if (tem == tt.substr(1, len))
		return true;
	return false;
}
bool is_first(string tem, string tt) {
	int len = tem.size();
	if (tem == tt.substr(0, len))
		return true;
	return false;
}
int main()
{
	cin >> n;
	int m = 2 * n - 2;
	for (int i = 1; i <= m; i++) {
		cin >> s[i];
		p[i].num = i;
		p[i].t = s[i];
		c[i] = 'S';
	}
	if (n == 2) {
		cout << "PS" << endl;
		return 0;
	}
	sort(p + 1, p + 1 + m, cmp);

	string tem1=p[1].t;
	string tem2 = p[2].t;
	c[p[1].num] = 'P';
	bool flag = 0;
	for (int i = 3; i <= m; i++) {
		if ((is_first(tem1,p[i].t)&&is_last(tem2,p[i+1].t))){ 
			tem1 = p[i].t;
			tem2 = p[i + 1].t;
			c[p[i].num] = 'P';
			i++;
		}
		else if (is_first(tem1, p[i + 1].t) && is_last(tem2, p[i].t)) {
			tem1 = p[i + 1].t;
			tem2 = p[i].t;
			c[p[i + 1].num] = 'P';
			i++;
		}
		else {
			flag = 1;
			for (int i = 1; i <= m; i++) {
				c[i] = 'S';
			}
			break;
		}	
	}
	
	if(flag==1) {
		tem1 = p[2].t;
		tem2 = p[1].t;
		c[p[2].num] = 'P';
		for (int i = 3; i <= m; i++) {
			if ((is_first(tem1, p[i].t) && is_last(tem2, p[i + 1].t))) {
				tem1 = p[i].t;
				tem2 = p[i + 1].t;
				c[p[i].num] = 'P';
				i++;
			}
			else {
				tem1 = p[i + 1].t;
				tem2 = p[i].t;
				c[p[i + 1].num] = 'P';
				i++;
			}
		}
	}
	for (int i = 1; i <= m; i++) 
		cout << c[i];
	cout << endl;
	return 0;
}