#include <iostream>
#include <vector>
#include <queue>
#include <set>

#define MAX_V 200010
using namespace std;

vector<int> G[MAX_V];
queue <int> que;
int a[MAX_V],p[MAX_V],n;

//サイクルを一周して最後に矛盾がないかどうか
bool cycle(int x, int y){
	int i = p[x];
	int pre = y;
	while(i!=x){
		set<int> s;
		s.insert(pre);
		for(int j=0;j<G[i].size();j++){
			s.insert(a[G[i][j]]);
		}
		int z = 0;
		while(s.count(z)){
			z++;
		}
		a[i] = z;
		pre = a[i];
		i = p[i];
	}
	set<int> st;
	st.insert(pre);
	for(int j=0;j<G[x].size();j++){
		st.insert(a[G[x][j]]);
	}
	int z = 0;
	while(st.count(z)){
		z++;
	}
	return y==z;
}

int main(){
	int i,j,z;
	cin >> n;
	for(i=0;i<n;i++){
		cin >> p[i];
		p[i]--;
		G[p[i]].push_back(i);
	}

	for(i=0;i<n;i++){
		if(G[i].size()==0){
			que.push(i);
		}
		a[i]  = -1;
	}

	while(!que.empty()){
		set<int> s;
		int x = que.front();
		que.pop();
		if(a[x]>=0) continue;
		for(j=0;j<G[x].size();j++){
			if(a[G[x][j]]<0){
				z = 1;
			}
			s.insert(a[G[x][j]]);
		}
		if(z){
			z = 0;
			continue;
		}
		int y = 0;
		while(s.count(y)){
			y++;
		}
		a[x] = y;
		que.push(p[x]);
	}

	for(i=0;i<n;i++){
		if(a[i]<0){
			z = i;
			break;
		}
	}
	
	set<int> st;
	for(i=0;i<G[z].size();i++){
		st.insert(a[G[z][i]]);
	}
	j = 0;
	while(st.count(j)){
		j++;
	}
	int k = j;
	j++;
	while(st.count(j)){
		j++;
	}
	int l = j;
	if(cycle(z,k)||cycle(z,l)){
		cout << "POSSIBLE" << endl;
		return 0;
	}
	cout << "IMPOSSIBLE" << endl;
}

### 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 <iostream>
#include <vector>
#include <queue>
#include <set>

#define MAX_V 200010
using namespace std;

vector<int> G[MAX_V];
queue <int> que;
int a[MAX_V],p[MAX_V],n;

//サイクルを一周して最後に矛盾がないかどうか
bool cycle(int x, int y){
	int i = p[x];
	int pre = y;
	while(i!=x){
		set<int> s;
		s.insert(pre);
		for(int j=0;j<G[i].size();j++){
			s.insert(a[G[i][j]]);
		}
		int z = 0;
		while(s.count(z)){
			z++;
		}
		a[i] = z;
		pre = a[i];
		i = p[i];
	}
	set<int> st;
	st.insert(pre);
	for(int j=0;j<G[x].size();j++){
		st.insert(a[G[x][j]]);
	}
	int z = 0;
	while(st.count(z)){
		z++;
	}
	return y==z;
}

int main(){
	int i,j,z;
	cin >> n;
	for(i=0;i<n;i++){
		cin >> p[i];
		p[i]--;
		G[p[i]].push_back(i);
	}

	for(i=0;i<n;i++){
		if(G[i].size()==0){
			que.push(i);
		}
		a[i]  = -1;
	}

	while(!que.empty()){
		set<int> s;
		int x = que.front();
		que.pop();
		if(a[x]>=0) continue;
		for(j=0;j<G[x].size();j++){
			if(a[G[x][j]]<0){
				z = 1;
			}
			s.insert(a[G[x][j]]);
		}
		if(z){
			z = 0;
			continue;
		}
		int y = 0;
		while(s.count(y)){
			y++;
		}
		a[x] = y;
		que.push(p[x]);
	}

	for(i=0;i<n;i++){
		if(a[i]<0){
			z = i;
			break;
		}
	}
	
	set<int> st;
	for(i=0;i<G[z].size();i++){
		st.insert(a[G[z][i]]);
	}
	j = 0;
	while(st.count(j)){
		j++;
	}
	int k = j;
	j++;
	while(st.count(j)){
		j++;
	}
	int l = j;
	if(cycle(z,k)||cycle(z,l)){
		cout << "POSSIBLE" << endl;
		return 0;
	}
	cout << "IMPOSSIBLE" << endl;
}

```