CodeForces 599B Spongebob and Joke【思维】

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.

这题暴力识别会TLE啊……OTZ

还是巧妙地用一下数组。为F的角标开一个新数组，新数组的下标是F的值，数组内容是F的角标。这样就可以直接输出Ai了。

然后判定是否可能这一点，写两个变量。一个记录没有重复的数，一个记录重复的。这两个量不能合成一个，成一个的话没法判断。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define Max 100011
using namespace std;
int a[Max],b[Max];
int fn[Max];
struct Fn
{
int num,n;
}f[Max];
int main()
{
int n,m,i,j;
while(~scanf("%d%d",&n,&m))
{
memset(fn,0,sizeof(fn));
int sum=0,temp=0;
for(i=1;i<=n;i++)
{
scanf("%d",&f[i].num);
f[f[i].num].n=i;
fn[f[i].num]++;
}
for(i=1;i<=m;i++)
{
scanf("%d",&b[i]);
if(fn[b[i]]==1)
sum++;
if(fn[b[i]]>1)
temp++;
}
if(sum+temp<m)
printf("Impossible\n");
else if(temp>0)
printf("Ambiguity\n");
else
{
int t=0;
printf("Possible\n");
for(i=1;i<=m;i++)
{
if(fn[b[i]]==1)
{
if(t>0)
printf(" ");
t++;
printf("%d",f[b[i]].n);
}
}
printf("\n");
}
}
return 0;
}