Spongebob and Joke（记录数据地址）

                         Spongebob and Joke

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatus

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 sequencea₁, a₂, ..., a_m of lengthm, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequencef₁, f₂, ..., f_n of lengthn and for each number a_i got number b_i = f_ai. To finish the prank he erased the initial sequencea_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 integersn andm (1 ≤ n, m ≤ 100 000) — the lengths of sequencesf_i andb_i respectively.

The second line contains n integers, determining sequencef₁, f₂, ..., f_n (1 ≤ f_i ≤ n).

The last line contains m integers, determining sequenceb₁, b₂, ..., b_m(1 ≤ b_i ≤ n).

Output

Print "Possible" if there is exactly one sequencea_i, such thatb_i = f_ai for alli from 1 tom. Then print m integersa₁, 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 sequencea_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 by1 and vice versa, while2 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 alli, so no sequencea_i transforms into suchb_i and we can say for sure that Spongebob has made a mistake.

给出两个集合，f与d，问是否存在另外一个集合a，使，f[a[i]]=b[i],这两天自学容器，初次试手

#include<stdio.h>#include<algorithm>#include<iostream>#include<map>using namespace std;int f[100005],b[100005];int main(){map<int,int>k1;//定义map容器map<int,int>k2;k1.clear();k2.clear();int n,m,x;int sum=0,num=0;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&f[i]);k1[f[i]]++;//键值k2[f[i]]=i;}//记录位置for(int i=1;i<=m;i++){scanf("%d",&b[i]);if(k1[b[i]]==1)sum++;if(k2[b[i]]==0)num++;}if(num!=0)printf("Impossible\n");if(num==0&&sum<m)printf("Ambiguity\n");if(sum==m){printf("Possible\n");for(int i=1;i<=m;i++){if(k1[b[i]]==1)printf("%d ",k2[b[i]]);}printf("\n");}}