Codeforces Round #332 (Div. 二) B. Spongebob and Joke

Description

While Patrick was gone shopping, Spongebob decided to play a little trick on his friend. The naughty Sponge browsed through Patrick‘s personal stuff and found a sequence a₁, a₂, ..., a_m of length m, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequence f₁, f₂, ..., f_n of length n and for each number a_i got number b_i = f_ai. To finish the prank he erased the initial sequence a_i.

It‘s hard to express how sad Patrick was when he returned home from shopping! We will just say that Spongebob immediately got really sorry about what he has done and he is now trying to restore the original sequence. Help him do this or determine that this is impossible.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100 000) — the lengths of sequences f_i and b_i respectively.

The second line contains n integers, determining sequence f₁, f₂, ..., f_n (1 ≤ f_i ≤ n).

The last line contains m integers, determining sequence b₁, b₂, ..., b_m(1 ≤ b_i ≤ n).

Output

Print "Possible" if there is exactly one sequence a_i, such that b_i = f_ai for all i from 1 to m. Then print m integers a₁, a₂, ..., a_m.

If there are multiple suitable sequences a_i, print "Ambiguity".

If Spongebob has made a mistake in his calculations and no suitable sequence a_i exists, print "Impossible".

Sample Input

Input

3 33 2 11 2 3

Output

Possible3 2 1 

Input

3 31 1 11 1 1

Output

Ambiguity

Input

3 31 2 13 3 3

Output

Impossible

Hint

In the first sample 3 is replaced by 1 and vice versa, while 2 never changes. The answer exists and is unique.

In the second sample all numbers are replaced by 1, so it is impossible to unambiguously restore the original sequence.

In the third sample f_i ≠ 3 for all i, so no sequence a_i transforms into such b_i and we can say for sure that Spongebob has made a mistake.

题意：给你数组f[]和数组b[]问你是否存在数组a[]使得bi==fai;

题解：三种情况1、数组b中的数字数组f中没有 则输出Impossible     2、数组b中的数字在数组f中重复出现且数组b中的数字都在数组f中输出Ambiguity      3、数组b中的数字都在数组f中且无重复输出Possible以及数组a[]

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define LL long long
#define MAX 100100
using namespace std;
int b[MAX],f[MAX];
int a[MAX],vis[MAX];
int main()
{
	int n,m,i,j,ans;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&ans);
			f[ans]=i;
			vis[ans]++;
		}
		for(i=1;i<=m;i++)
			scanf("%d",&b[i]);
		int flag=0;
		int Flag=0;
		for(i=1;i<=m;i++)
		{
			if(vis[b[i]]==0)
			{
				flag=1;
				break;
			}
			if(vis[b[i]]>1)
			    Flag=1;
		}
		if(flag)
		{
			printf("Impossible\n");
			continue;
		}
		if(Flag)
		{
			printf("Ambiguity\n");
			continue;
		}
		for(i=1;i<=m;i++)
			a[i]=f[b[i]];
		printf("Possible\n");
		for(i=1;i<m;i++)
		printf("%d ",a[i]);
		printf("%d\n",a[m]);
	}
	return 0;
}