Codeforces Round #332 (Div. 2) B. Spongebob and Joke(水题，构造)

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 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.

【题意】给你n个f[i],m个b[i]，然后问你能不能找到m个a[i]，使得b[i]=f[a[i]].

【解题方法】简单构造。暴力存一下这个数在f[i]中出现了多少次，如果没出现就输出impossilbe，如果出现多次，就多解，如果出现一次就输出这个数。

【AC 代码】

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int f[maxn],b[maxn],a[maxn];
vector<int>G[maxn];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++) scanf("%d",&f[i]);
for(int i=1; i<=m; i++) scanf("%d",&b[i]);
for(int i=1; i<=n; i++){
G[f[i]].push_back(i);
}
bool ok=0;
for(int i=1; i<=m; i++){
if(G[b[i]].size()==0){
puts("Impossible");
return 0;
}
if(G[b[i]].size()>1) ok=1;
}
if(ok==1){
puts("Ambiguity");
return 0;
}
puts("Possible");
for(int i=1; i<=m; i++){
printf("%d ",G[b[i]][0]);
}
return 0;
}