【Codeforces】-599B-Spongebob and Joke（思维）

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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".

Examples

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.

这道题真的好烦，一个下午用几种方法写，写到一半都觉得不能继续，然后我就废了！

1  记录a中每个数出现的序号，和这个数出现的次数。再判断b里的数是不是都在a中出现过，不是就是不可能。全部出现且出现不止一次就是不确定的。

2  如果都只出现一次，依次输出与b相等的a中数的序号。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{int num,n;}a[1000010];int main(){int n,m;int b[100010];int ant[100010]={0};int num1,num2;while(~scanf("%d %d",&n,&m)){num1=0;num2=0;memset(ant,0,sizeof(ant));for(int i=1;i<=n;i++){scanf("%d",&a[i].num);a[a[i].num].n=i;//记录这个数的序号 ant[a[i].num]++;//记录每个数字出现的次数 }for(int i=1;i<=m;i++){scanf("%d",&b[i]);if(ant[b[i]]==1)//num1，num2计数，没有出现时ant[i]==0不记下来 num1++;else if(ant[b[i]]>1)num2++;}if(num1+num2<m)//成立说明存在一个数在 f[i]中没有出现过，就是不可能 printf("Impossible\n");else {if(num2>0)printf("Ambiguity\n");//一个数出现不止一次，就是不确定的 else{printf("Possible\n");int z=m;for(int i=1;i<=m;i++){if(ant[b[i]]==1)//扫b[i]这个序列，出现一次输出a中与b相等值的序号 printf("%d",a[b[i]].n);if(z>0){printf(" ");z--;}}printf("\n");}}}return 0;}