#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int N=2e5+11;
queue<int> Q;
vector<int> V[N];
int vis[N],tmp[N],from[N],d[N],n,pre,ans,tim,col[N],i,j,k,u;
int main(){
//	freopen("ex.in","r",stdin);
	scanf("%d",&n);
	for(i=1;i<=n;++i){
		scanf("%d",&from[i]);
		d[from[i]]++;
	}
	for(i=1;i<=n;++i) if(d[i]==0) Q.push(i);
	while(!Q.empty()){
		u=Q.front(),Q.pop();
		tim++;
		for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
		for(i=0;tmp[i]==tim;++i);
		col[u]=i;
		V[from[u]].push_back(i);
		if(!(--d[from[u]])) Q.push(from[u]);
	}
	for(i=1;i<=n;++i) if(d[i]){u=i;break;}
	tim++;
	for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
	for(k=0;tmp[k]==tim;++k);
	ans=pre=k;
	for(i=from[u];i!=u;i=from[i]){
		tim++;
		tmp[pre]=tim;
		for(j=0;j<V[i].size();++j) tmp[V[i][j]]=tim;
		for(j=0;tmp[j]==tim;++j);
		pre=j;
	}
	if(pre!=ans){
		puts("POSSIBLE");
		return 0;
	}
	tim++;
	for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
	for(k=0;tmp[k]==tim;++k);
	ans=k;
	for(k++;tmp[k]==tim;++k);
	pre=k;
	for(i=from[u];i!=u;i=from[i]){
		tim++;
		tmp[pre]=tim;
		for(j=0;j<V[i].size();++j) tmp[V[i][j]]=tim;
		for(j=0;tmp[j]==tim;++j);
		pre=j;
	}
	if(pre==ans) puts("POSSIBLE");
	else puts("IMPOSSIBLE");
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int N=2e5+11;
queue<int> Q;
vector<int> V[N];
int vis[N],tmp[N],from[N],d[N],n,pre,ans,tim,col[N],i,j,k,u;
int main(){
//	freopen("ex.in","r",stdin);
	scanf("%d",&n);
	for(i=1;i<=n;++i){
		scanf("%d",&from[i]);
		d[from[i]]++;
	}
	for(i=1;i<=n;++i) if(d[i]==0) Q.push(i);
	while(!Q.empty()){
		u=Q.front(),Q.pop();
		tim++;
		for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
		for(i=0;tmp[i]==tim;++i);
		col[u]=i;
		V[from[u]].push_back(i);
		if(!(--d[from[u]])) Q.push(from[u]);
	}
	for(i=1;i<=n;++i) if(d[i]){u=i;break;}
	tim++;
	for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
	for(k=0;tmp[k]==tim;++k);
	ans=pre=k;
	for(i=from[u];i!=u;i=from[i]){
		tim++;
		tmp[pre]=tim;
		for(j=0;j<V[i].size();++j) tmp[V[i][j]]=tim;
		for(j=0;tmp[j]==tim;++j);
		pre=j;
	}
	if(pre!=ans){
		puts("POSSIBLE");
		return 0;
	}
	tim++;
	for(i=0;i<V[u].size();++i) tmp[V[u][i]]=tim;
	for(k=0;tmp[k]==tim;++k);
	ans=k;
	for(k++;tmp[k]==tim;++k);
	pre=k;
	for(i=from[u];i!=u;i=from[i]){
		tim++;
		tmp[pre]=tim;
		for(j=0;j<V[i].size();++j) tmp[V[i][j]]=tim;
		for(j=0;tmp[j]==tim;++j);
		pre=j;
	}
	if(pre==ans) puts("POSSIBLE");
	else puts("IMPOSSIBLE");
	return 0;
}
```