#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>

using namespace std;
const int MX=200011;
int n,p[MX];
int v[MX],nv[MX],inr[MX];
vector<int>G[MX];
typedef vector<int>::iterator vit;

void calv(int k){
	if(~v[k])return;
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])calv(*i);
	static int t[MX]={0};
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])++t[v[*i]];
	int I;
	for(I=0;I<n;I++)if(!t[I]){v[k]=I;break;}
	for(I++;I<n;I++)if(!t[I]){nv[k]=I;break;}
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])--t[v[*i]];
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&p[i]),G[p[i]].push_back(i);
	
	int x=1,y=1;
	do x=p[x],y=p[p[y]];
	while(x!=y);
	for(;!inr[x];inr[x]=1,x=p[x]);
	
	memset(v+1,-1,sizeof(int)*n);
	for(int i=1;i<=n;i++)calv(i);
	
	static int t[MX];
	memset(t+1,-1,sizeof(int)*n);
	t[p[x]]=v[p[x]];
	for(y=p[x];y!=x;y=p[y])t[p[y]]=(t[y]==v[p[y]]?nv[p[y]]:v[p[y]]);
	if(t[p[x]]==(t[x]==v[p[x]]?nv[p[x]]:v[p[x]]))return puts("POSSIBLE"),0;
	
	memset(t+1,-1,sizeof(int)*n);
	t[p[x]]=nv[p[x]];
	for(y=p[x];y!=x;y=p[y])t[p[y]]=(t[y]==v[p[y]]?nv[p[y]]:v[p[y]]);
	if(t[p[x]]==(t[x]==v[p[x]]?nv[p[x]]:v[p[x]]))return puts("POSSIBLE"),0;
	
	puts("IMPOSSIBLE");
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.

The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.

We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.

* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.



Determine whether there exists such an assignment.

Constraints

* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.

Examples

Input

4
2 3 4 1


Output

POSSIBLE


Input

3
2 3 1


Output

IMPOSSIBLE


Input

4
2 3 1 1


Output

POSSIBLE


Input

6
4 5 6 5 6 4


Output

IMPOSSIBLE

### Response

```cpp
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>

using namespace std;
const int MX=200011;
int n,p[MX];
int v[MX],nv[MX],inr[MX];
vector<int>G[MX];
typedef vector<int>::iterator vit;

void calv(int k){
	if(~v[k])return;
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])calv(*i);
	static int t[MX]={0};
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])++t[v[*i]];
	int I;
	for(I=0;I<n;I++)if(!t[I]){v[k]=I;break;}
	for(I++;I<n;I++)if(!t[I]){nv[k]=I;break;}
	for(vit i=G[k].begin();i!=G[k].end();i++)if(!inr[*i])--t[v[*i]];
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&p[i]),G[p[i]].push_back(i);
	
	int x=1,y=1;
	do x=p[x],y=p[p[y]];
	while(x!=y);
	for(;!inr[x];inr[x]=1,x=p[x]);
	
	memset(v+1,-1,sizeof(int)*n);
	for(int i=1;i<=n;i++)calv(i);
	
	static int t[MX];
	memset(t+1,-1,sizeof(int)*n);
	t[p[x]]=v[p[x]];
	for(y=p[x];y!=x;y=p[y])t[p[y]]=(t[y]==v[p[y]]?nv[p[y]]:v[p[y]]);
	if(t[p[x]]==(t[x]==v[p[x]]?nv[p[x]]:v[p[x]]))return puts("POSSIBLE"),0;
	
	memset(t+1,-1,sizeof(int)*n);
	t[p[x]]=nv[p[x]];
	for(y=p[x];y!=x;y=p[y])t[p[y]]=(t[y]==v[p[y]]?nv[p[y]]:v[p[y]]);
	if(t[p[x]]==(t[x]==v[p[x]]?nv[p[x]]:v[p[x]]))return puts("POSSIBLE"),0;
	
	puts("IMPOSSIBLE");
	return 0;
}

```