codeforces #332 div 2 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 integersa1, 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 3
3 2 1
1 2 3

output

Possible
3 2 1 

input

3 3
1 1 1
1 1 1

output

Ambiguity

input

3 3
1 2 1
3 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.

理解题意。。

需要注意的是。。。只有当在f中重复出现的那个在b中也出现了。。答案才是任意。。不然五影响。。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

const int N=1E5+7;
int n,m;
int ans[N];
bool v[N];
bool v2[N];
using namespace std;

struct node
{
    int val;
    int id;
}f[N],b[N];

bool cmp( node a,node b)
{
    return a.val<b.val;
}
int main()
{
    cin>>n>>m;
    memset(v2,false,sizeof(v2));
    for ( int i = 0 ; i < n ; i++) scanf("%d",&f[i].val),f[i].id=i;
    for ( int i = 0 ; i < m ; i++) scanf("%d",&b[i].val),b[i].id=i;
    for ( int i = 0 ; i < m ; i++) v2[b[i].val] = true;

    sort(f,f+n,cmp);
    sort(b,b+m,cmp);

    bool multi = false;
    memset(v,false,sizeof(v));
    for ( int i = 0 ; i < n-1 ; i++) if (f[i].val==f[i+1].val&&v2[f[i+1].val]) multi = true;
    for ( int i = 0 ; i < n ; i++) v[f[i].val] = true;
    bool sad = false;
    for ( int i = 0 ; i < m; i++) if (!v[b[i].val]) sad = true;
    memset(ans,0,sizeof(ans));
    int j = 0;
    bool flag = false;
    for ( int i = 0 ; i < n ; i++)
    {
        if (j==m) break;
        if (f[i].val==b[j].val&&j<m)
        {
            while (f[i].val==b[j].val&&j<m)
            {
                ans[b[j].id] = f[i].id;
                j++;
            }
        }

    }
     if (sad)
     {
         puts("Impossible");
     }
     else
     {
         if (multi) puts("Ambiguity");
         else
         {
             puts("Possible");
             for ( int i = 0 ; i < m-1 ; i++)
                printf("%d ",ans[i]+1);
                printf("%d
",ans[m-1]+1);
         }

     }




}