Codeforces 599B： Spongebob and Joke（技巧，规律）

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.

题目大意：

给你一个f序列，一个b序列，问是否能从f序列中抽出某些数来组成序列b，如果能的话输出可能，并输出是从f序列的第几个数中组成b序列中的第1个，第2个，第3个...数的，如果不能的话输出不可能，如果有多种情况的话输出模棱两可。

解题思路：

不可能的：下面有某个数，上面没有；不确定的：上面某个数出现了多次；其余的就是可能的情况了。按照这个顺序来判断。

代码如下：

#include <stdio.h>
struct node
{
int id,num;//id是第一行某个数是第几个出现的，num就是那个数。
}per[100010];//记录第一行
int b[100010];//记录第二行的数
int ci[100010];//记录第一行的数们出现的次数
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&per[i].num);
per[per[i].num].id=i;//技巧。。
ci[per[i].num]++;
}
int sum=0;
int flag2=0;//标记第二行中所出现的数在第一行中是否出现多次
int flag1=0;//标记 第二行是否存在一些数，而第一行没有的
for(int i=1;i<=m;i++)
{
scanf("%d",&b[i]);
if(ci[b[i]]==0)
{
flag1=1;
}
if(ci[b[i]]>1)
flag2=1;
}
if(flag1==1)
{
printf("Impossible\n");
return 0;
}
if(flag2==1)
{
printf("Ambiguity\n");
return 0;
}
printf("Possible\n");
int t=0;
for(int i=1;i<=m;i++)
{
if(t>0)
{
printf(" ");
}
t++;
printf("%d",per[b[i]].id);
}
printf("\n");
}