#include<cstdio>
#include<queue>
const int N=2e5+5;
int p[N],head[N],cnt,n,deg[N],val[N];
bool vis[N],ur[N];
struct edge{
	int to,nxt;
}e[N];
std::queue<int>q;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",p+i);
		e[++cnt]=(edge){i,head[p[i]]},head[p[i]]=cnt,++deg[p[i]];
	}
	for(int i=1;i<=n;++i)if(!deg[i])q.push(i);
	while(!q.empty()){
		int u=q.front();q.pop();
		vis[u]=1;
		for(int i=head[u];i;i=e[i].nxt)if(vis[e[i].to])
			ur[val[e[i].to]]=1;
		int&v=val[u];
		while(ur[v])++v;
		for(int i=head[u];i;i=e[i].nxt)ur[val[e[i].to]]=0;
		if(!--deg[p[u]])q.push(p[u]);
	}
	int len=0,mn=0;
	for(int u=1;u<=n;++u)if(!vis[u]){
		for(int i=head[u];i;i=e[i].nxt)if(vis[e[i].to])
			ur[val[e[i].to]]=1;
		int&v=val[u];
		while(ur[v])++v;
		for(int i=head[u];i;i=e[i].nxt)ur[val[e[i].to]]=0;
	}
	for(int i=1;i<=n;++i)if(!vis[i]){
		++len;if(val[p[i]]<val[i])mn=p[i];
	}
	printf("%sPOSSIBLE\n",(len&1)&&!mn?"IM":"");
	return 0;
}
### Prompt

Generate a CPP solution to the following problem:

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<queue>
const int N=2e5+5;
int p[N],head[N],cnt,n,deg[N],val[N];
bool vis[N],ur[N];
struct edge{
	int to,nxt;
}e[N];
std::queue<int>q;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",p+i);
		e[++cnt]=(edge){i,head[p[i]]},head[p[i]]=cnt,++deg[p[i]];
	}
	for(int i=1;i<=n;++i)if(!deg[i])q.push(i);
	while(!q.empty()){
		int u=q.front();q.pop();
		vis[u]=1;
		for(int i=head[u];i;i=e[i].nxt)if(vis[e[i].to])
			ur[val[e[i].to]]=1;
		int&v=val[u];
		while(ur[v])++v;
		for(int i=head[u];i;i=e[i].nxt)ur[val[e[i].to]]=0;
		if(!--deg[p[u]])q.push(p[u]);
	}
	int len=0,mn=0;
	for(int u=1;u<=n;++u)if(!vis[u]){
		for(int i=head[u];i;i=e[i].nxt)if(vis[e[i].to])
			ur[val[e[i].to]]=1;
		int&v=val[u];
		while(ur[v])++v;
		for(int i=head[u];i;i=e[i].nxt)ur[val[e[i].to]]=0;
	}
	for(int i=1;i<=n;++i)if(!vis[i]){
		++len;if(val[p[i]]<val[i])mn=p[i];
	}
	printf("%sPOSSIBLE\n",(len&1)&&!mn?"IM":"");
	return 0;
}
```