#include<algorithm>
#include<iostream>
#include<queue>
#include<set>
using namespace std;
const int N=2e5+5;
set<int>se[N];
typedef set<int>::iterator it;
int n,now,v[N],c[N],b[N],nex[N],head[N];
struct each{
	int a,id;
}a[N];
queue<int>que;
inline int read(){
	register int x=0;
	register char ch=getchar();
	while(ch<'0'||ch>'9')	ch=getchar();
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x;
}
int get(int x){
	int last=-1;
	it i=se[x].begin();
	for(;i!=se[x].end();++i)
		if(last+1==*i)	++last;
		else	break;
	return ++last;
}
int vis[N],tm[N],p[N];
int main(){
	//freopen("graph.out","w",stdout);
	//freopen("graph.in","r",stdin);

	n=read();
	for(register int i=1;i<=n;i++)
		++b[p[i]=read()];
	for(register int i=1;i<=n;i++)
		if(!b[i])	que.push(i);
	while(!que.empty()){
		int x=que.front(),k=get(x);
		que.pop(),se[p[x]].insert(k);
		if(!(--b[p[x]]))	que.push(p[x]);
	}
	int st;
	for(int i=1;i<=n;i++)
		if(b[i]){
			st=i,que.push(i);
			break;
		}
	while(!que.empty()){
		int x=que.front();
		que.pop(),tm[x]=get(x);
		if(se[p[x]].count(tm[x]))	vis[x]=1;
		se[p[x]].insert(tm[x]);
		if(p[x]==st){
			if(tm[x]==tm[st])	break;
			else{
				puts("POSSIBLE");
				return 0;
			}
		}
		que.push(p[x]);
	}
	for(int i=1;i<=n;i++)
		if(b[i]&&!vis[i])	se[p[i]].erase(tm[i]);
	int end=tm[st];
	que.push(st),se[st].insert(end);
	while(!que.empty()){
		int x=que.front();
		que.pop(),tm[x]=get(x);
		se[p[x]].insert(tm[x]);
		if(p[x]==st){
			if(tm[x]!=end)	break;
			else{
				puts("POSSIBLE");
				return 0;
			}
		}
		que.push(p[x]);
	}
	puts("IMPOSSIBLE");
	return 0;
}
### Prompt

Please formulate 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<algorithm>
#include<iostream>
#include<queue>
#include<set>
using namespace std;
const int N=2e5+5;
set<int>se[N];
typedef set<int>::iterator it;
int n,now,v[N],c[N],b[N],nex[N],head[N];
struct each{
	int a,id;
}a[N];
queue<int>que;
inline int read(){
	register int x=0;
	register char ch=getchar();
	while(ch<'0'||ch>'9')	ch=getchar();
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x;
}
int get(int x){
	int last=-1;
	it i=se[x].begin();
	for(;i!=se[x].end();++i)
		if(last+1==*i)	++last;
		else	break;
	return ++last;
}
int vis[N],tm[N],p[N];
int main(){
	//freopen("graph.out","w",stdout);
	//freopen("graph.in","r",stdin);

	n=read();
	for(register int i=1;i<=n;i++)
		++b[p[i]=read()];
	for(register int i=1;i<=n;i++)
		if(!b[i])	que.push(i);
	while(!que.empty()){
		int x=que.front(),k=get(x);
		que.pop(),se[p[x]].insert(k);
		if(!(--b[p[x]]))	que.push(p[x]);
	}
	int st;
	for(int i=1;i<=n;i++)
		if(b[i]){
			st=i,que.push(i);
			break;
		}
	while(!que.empty()){
		int x=que.front();
		que.pop(),tm[x]=get(x);
		if(se[p[x]].count(tm[x]))	vis[x]=1;
		se[p[x]].insert(tm[x]);
		if(p[x]==st){
			if(tm[x]==tm[st])	break;
			else{
				puts("POSSIBLE");
				return 0;
			}
		}
		que.push(p[x]);
	}
	for(int i=1;i<=n;i++)
		if(b[i]&&!vis[i])	se[p[i]].erase(tm[i]);
	int end=tm[st];
	que.push(st),se[st].insert(end);
	while(!que.empty()){
		int x=que.front();
		que.pop(),tm[x]=get(x);
		se[p[x]].insert(tm[x]);
		if(p[x]==st){
			if(tm[x]!=end)	break;
			else{
				puts("POSSIBLE");
				return 0;
			}
		}
		que.push(p[x]);
	}
	puts("IMPOSSIBLE");
	return 0;
}
```