Chouti and his classmates are going to the university soon. To say goodbye to each other, the class has planned a big farewell party in which classmates, teachers and parents sang and danced.
Chouti remembered that n persons took part in that party. To make the party funnier, each person wore one hat among n kinds of weird hats numbered 1,2,…n. It is possible that several persons wore hats of the same kind. Some kinds of hats can remain unclaimed by anyone.
After the party, the i-th person said that there were ai persons wearing a hat differing from his own.
It has been some days, so Chouti forgot all about others’ hats, but he is curious about that. Let bi be the number of hat type the i-th person was wearing, Chouti wants you to find any possible b1,b2,…,bn that doesn’t contradict with any person’s statement. Because some persons might have a poor memory, there could be no solution at all.
Input
The first line contains a single integer n (1≤n≤105), the number of persons in the party.
The second line contains n integers a1,a2,…,an (0≤ai≤n−1), the statements of people.
Output
If there is no solution, print a single line “Impossible”.
Otherwise, print “Possible” and then n integers b1,b2,…,bn (1≤bi≤n).
If there are multiple answers, print any of them.
Examples
Input
3
0 0 0
Output
Possible
1 1 1
Input
5
3 3 2 2 2
Output
Possible
1 1 2 2 2
Input
4
0 1 2 3
Output
Impossible
Note
In the answer to the first example, all hats are the same, so every person will say that there were no persons wearing a hat different from kind 1.
In the answer to the second example, the first and the second person wore the hat with type 1 and all other wore a hat of type 2.
So the first two persons will say there were three persons with hats differing from their own. Similarly, three last persons will say there were two persons wearing a hat different from their own.
In the third example, it can be shown that no solution exists.
In the first and the second example, other possible configurations are possible.
题意：有n个人，每个人都戴了一顶帽子，有些人的帽子颜色是一样的，每个人都知道有多少人帽子颜色与自己不一样，都是呢，有些人记性不好，忘记了具体有多少人与自己不同，现在告诉你每个人的表述（有多少个人与自己帽子颜色不一样），问你可不可以还原原来每个帽子的颜色，如果可以输出可能，接下来一个序列代表帽子颜色，否则输出不可能
思路：开始使劲怼正面，几个小时都没有做出来，然后分析一下反面，告诉我有多少个人与自己不同，那么用总数n减去a[i]就是和自己帽子颜色相同的人的个数，所以统计n-a[i]的个数就可以了
如果不能取余就不可能凑出来。
#include<bits/stdc++.h>
#define LL long long
#define Max 100005
#define Mod 1e9+7
const LL mod=1e9+7;
const LL inf=0x3f3f3f3f;
using namespace std;
int n,num[Max],ans[Max];
struct node{
int data,index;
}a[Max];//方便等下构造序列
bool cmp(node a,node b){
return a.data<b.data;
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i].data);
a[i].data=n-a[i].data;
num[a[i].data]++;//统计
a[i].index=i;
}
sort(a,a+n,cmp);
for(int i=0;i<n;i++){
if(num[a[i].data]%a[i].data!=0){//不能取余，直接输出不可能
printf("Impossible\n");
return 0;
}
}
int t=1;
for(int i=0;i<n;i++){
if((a[i].data==a[i-1].data && num[a[i].data]%a[i].data==0))//特殊情况，
//帽子颜色不同但是不同的人数一样，如aaabbcc，所以帽子颜色要+1
t++;
ans[a[i].index]=t;//因为是个序列，不能改变顺序，之前排序了，所以要在原来下标位置赋值
num[a[i].data]--;
if(num[a[i].data]==0)
t++;//同一种颜色的帽子构造完了，颜色+1
}
printf("Possible\n");
for(int i=0;i<n;i++)
printf("%d ",ans[i]);
return 0;
}