#include <iostream>
#include <cstdio>
#include <cstring>
#define N 200005
using namespace std;
struct edge{
	int k,next;
}e[N<<1];
int n,home[N],cnt=-1,h[N],a[N],tot;
void add(int x,int y){
	cnt++;
	e[cnt].k=y;
	e[cnt].next=home[x];
	home[x]=cnt;
}
int s[N],top,nxt[N],vis[N],D;
void dfs(int k){
	vis[k]=D;s[++top]=k;
	for(int i=home[k];~i;i=e[i].next){
		if(!vis[e[i].k]) dfs(e[i].k);
		else if(vis[e[i].k]==D){
			tot=top;
			for(int p=1;p<=top;p++) h[s[p]]=s[p==1?top:p-1],nxt[s[p]]=s[p==top?1:p+1],a[p]=s[p];
		}
	}
	top--;
}
int f[N],sum[25];
void dfs(int k,int F){
	for(int i=home[k];~i;i=e[i].next) if(e[i].k!=F){
		dfs(e[i].k,k);
	}
	for(int i=home[k];~i;i=e[i].next) if(e[i].k!=F){
		sum[f[e[i].k]]=k;
	}
	for(int i=0;i<=20;i++) if(sum[i]!=k){
		f[k]=i;break;
	}
}
bool pan(){
	for(int i=f[a[1]];i<=20;i++){
		memset(sum,0,sizeof(sum));
		bool t=1;
		f[a[1]]=i;
		for(int p=home[a[1]];~p;p=e[p].next) if(e[p].k!=nxt[a[1]]) if(f[e[p].k]==i){
			t=0;break;
		}
		if(!t) continue;
		for(int p=tot;p;p--){
			for(int j=home[a[p]];~j;j=e[j].next) sum[f[e[j].k]]=p;
			for(int j=0;j<=20;j++) if(sum[j]!=p){
				f[a[p]]=j;break;
			}
		}
		if(f[a[1]]!=i) t=0;
		if(t) return 1;
	}
	return 0;
}
int main(){
	memset(home,-1,sizeof(home));
	scanf("%d",&n);
	for(int i=1,x;i<=n;i++){
		scanf("%d",&x);
		add(x,i);
	}
	for(int i=1;i<=n;i++) if(!vis[i]) D=i,dfs(i);
	for(int i=1;i<=n;i++) if(nxt[i]) dfs(i,nxt[i]);
	if(pan()) puts("POSSIBLE");
	else puts("IMPOSSIBLE");
	return 0;
}
### Prompt

Your challenge is to write 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 <iostream>
#include <cstdio>
#include <cstring>
#define N 200005
using namespace std;
struct edge{
	int k,next;
}e[N<<1];
int n,home[N],cnt=-1,h[N],a[N],tot;
void add(int x,int y){
	cnt++;
	e[cnt].k=y;
	e[cnt].next=home[x];
	home[x]=cnt;
}
int s[N],top,nxt[N],vis[N],D;
void dfs(int k){
	vis[k]=D;s[++top]=k;
	for(int i=home[k];~i;i=e[i].next){
		if(!vis[e[i].k]) dfs(e[i].k);
		else if(vis[e[i].k]==D){
			tot=top;
			for(int p=1;p<=top;p++) h[s[p]]=s[p==1?top:p-1],nxt[s[p]]=s[p==top?1:p+1],a[p]=s[p];
		}
	}
	top--;
}
int f[N],sum[25];
void dfs(int k,int F){
	for(int i=home[k];~i;i=e[i].next) if(e[i].k!=F){
		dfs(e[i].k,k);
	}
	for(int i=home[k];~i;i=e[i].next) if(e[i].k!=F){
		sum[f[e[i].k]]=k;
	}
	for(int i=0;i<=20;i++) if(sum[i]!=k){
		f[k]=i;break;
	}
}
bool pan(){
	for(int i=f[a[1]];i<=20;i++){
		memset(sum,0,sizeof(sum));
		bool t=1;
		f[a[1]]=i;
		for(int p=home[a[1]];~p;p=e[p].next) if(e[p].k!=nxt[a[1]]) if(f[e[p].k]==i){
			t=0;break;
		}
		if(!t) continue;
		for(int p=tot;p;p--){
			for(int j=home[a[p]];~j;j=e[j].next) sum[f[e[j].k]]=p;
			for(int j=0;j<=20;j++) if(sum[j]!=p){
				f[a[p]]=j;break;
			}
		}
		if(f[a[1]]!=i) t=0;
		if(t) return 1;
	}
	return 0;
}
int main(){
	memset(home,-1,sizeof(home));
	scanf("%d",&n);
	for(int i=1,x;i<=n;i++){
		scanf("%d",&x);
		add(x,i);
	}
	for(int i=1;i<=n;i++) if(!vis[i]) D=i,dfs(i);
	for(int i=1;i<=n;i++) if(nxt[i]) dfs(i,nxt[i]);
	if(pan()) puts("POSSIBLE");
	else puts("IMPOSSIBLE");
	return 0;
}
```