codeforce 599 B Spongebob and Joke

B. Spongebob and Joke

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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, ..., am of length m, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequence f1, f2, ..., fn of length n and for each number ai got number bi = fai. To finish the prank he erased the initial sequence ai.

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 fi and bi respectively.

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

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

Output

Print "Possible" if there is exactly one sequence ai, such that bi = fai for all i from 1 to m. Then print m integers a1, a2, ..., am.

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

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

Sample test(s)

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

Note

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 fi ≠ 3 for all i, so no sequence ai transforms into such bi and we can say for sure that Spongebob has made a mistake.

题意：嗯，现在懂了，这题是求f[a[i]]=b[i] ，多对一映射关系，求a[i]是下标，位置

Impossible在
Ambiguity之前

#include <cstdio>#include <cstring> int cnt[100000+10];int pos[100000+10];int b[100000+10];int main(){int n,m,a;while(~scanf("%d%d",&n,&m)){memset(cnt,0,sizeof(cnt));memset(pos,0,sizeof(pos));for(int i=1;i<=n;i++){scanf("%d",&a);cnt[a]++;pos[a]=i;}int f=0,f1=0;for(int i=1;i<=m;i++){scanf("%d",&b[i]);if(cnt[b[i]]==0)//b[i]在f中没有出现f=1;if(cnt[b[i]]>1)//b[i]在f中出现多次f1=1;}if(f==0&&f1==0){printf("Possible\n");printf("%d",pos[b[1]]);for(int i=2;i<=m;i++){printf(" %d",pos[b[i]]);}printf("\n");}else if(f)printf("Impossible\n");elseprintf("Ambiguity\n");}return 0;}