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.

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).

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".

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.

打了好几8次cf了 一直掉到灰名  昨天终于上分了 绿了 关键的一题（纪念*半夜三更打cf的日子）

题意： 输入 n,m

第二行 n个整数  f₁, f₂, ..., f_n (1 ≤ f_i ≤ n).

第三行 m个整数  b₁, b₂, ..., b_m (1 ≤ b_i ≤ n).

判断 有且只有一个b_i = f_ai 输出 fai的位置；

当b 没有在 f出现时输出 Impossible

当b 在f中出现 但f的位置不唯一时输出Ambiguity

除此之外 输出 Possible 及b1...bm 对应的位置  看懂题意 代码很简单

比赛的时候 被hack了一次

早点睡了 c题要补

#include<bits/stdc++.h>

using namespace std;

map<int,int>mp1;

map<int,int>mp2;

map<int,int>mp3;

int n,m;

int a,b;

int s[100005];

int re[100005];

int main()

{

    mp1.clear();

    mp2.clear();

    mp3.clear();

    memset(s,0,sizeof(s));

    memset(re,0,sizeof(re));

    scanf("%d%d",&n,&m);

    for(int i=0;i<n;i++)

       {

         scanf("%d",&a);

         mp1[a]++;

         mp2[a]=i+1;

       }

    for(int i=0;i<m;i++)

       {

         scanf("%d",&s[i]);

         mp3[s[i]]=1;

       }

     for(int i=0;i<m;i++)

       {

           if(mp2[s[i]]==0)

           {

               printf("Impossible\n");

               return 0;

           }

           re[i]=mp2[s[i]];

       }

       for(int i=1;i<=n;i++)

           {

               if(mp1[i]>=2&&mp3[i])

               {

                   printf("Ambiguity\n");

                   return 0;

               }

           }

           printf("Possible\n");

           printf("%d",re[0]);

           for(int i=1;i<m;i++)

           printf(" %d",re[i]);

           printf("\n");

    return 0;

}