#include <iostream>
#include <vector>
#include <set>
using namespace std;

vector<int> G[200000],cycle;
int used[200000],a[200000],nxt;

bool dfs(int v,int par){
	for(int to : G[v]){
		if(used[to] == par){
			cycle.push_back(v);
			return true;
		}else if(used[to] != -1) continue;
		used[to] = v;
		if(dfs(to,par)){
			cycle.push_back(v);
			return true;
		}
	}
	return false;
}
int calc(int v,int ext){
	if(a[v] != -1) return a[v];
	set<int> st;
	for(int to : G[v]){
		if(to == ext) continue;
		st.insert(calc(to,-1));
	}
	a[v] = 0;
	while(st.count(a[v])) a[v]++;
	if(v == cycle[0]){
		nxt = a[v] + 1;
		while(st.count(nxt)) nxt++;
	}
	return a[v];
}

int main(){
	int n,p[200000];
	cin >> n;
	for(int i = 0;i < n;i++){
		cin >> p[i];p[i]--;
		G[p[i]].push_back(i);
		used[i] = -1;
		a[i] = -1;
	}
	for(int i = 0;i < n;i++){
		if(used[i] == -1) {
			used[i] = i;
			dfs(i,i);
		}
	}
	bool flag = false;
	for(int i = 0;i < 2;i++){
		calc(cycle[0],cycle[cycle.size() - 1]);
		if(i) a[cycle[0]] = nxt;
		for(int j = 1;j < cycle.size();j++) calc(cycle[j],-1);
		int tmp = a[cycle[0]];
		a[cycle[0]] = -1;
		calc(cycle[0],-1);
		if(tmp == a[cycle[0]]) flag = true;
		for(int j = 0;j < n;j++) a[j] = -1;
	}
	if(flag) cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
	return 0;
}
### Prompt

Create a solution in CPP for 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 <vector>
#include <set>
using namespace std;

vector<int> G[200000],cycle;
int used[200000],a[200000],nxt;

bool dfs(int v,int par){
	for(int to : G[v]){
		if(used[to] == par){
			cycle.push_back(v);
			return true;
		}else if(used[to] != -1) continue;
		used[to] = v;
		if(dfs(to,par)){
			cycle.push_back(v);
			return true;
		}
	}
	return false;
}
int calc(int v,int ext){
	if(a[v] != -1) return a[v];
	set<int> st;
	for(int to : G[v]){
		if(to == ext) continue;
		st.insert(calc(to,-1));
	}
	a[v] = 0;
	while(st.count(a[v])) a[v]++;
	if(v == cycle[0]){
		nxt = a[v] + 1;
		while(st.count(nxt)) nxt++;
	}
	return a[v];
}

int main(){
	int n,p[200000];
	cin >> n;
	for(int i = 0;i < n;i++){
		cin >> p[i];p[i]--;
		G[p[i]].push_back(i);
		used[i] = -1;
		a[i] = -1;
	}
	for(int i = 0;i < n;i++){
		if(used[i] == -1) {
			used[i] = i;
			dfs(i,i);
		}
	}
	bool flag = false;
	for(int i = 0;i < 2;i++){
		calc(cycle[0],cycle[cycle.size() - 1]);
		if(i) a[cycle[0]] = nxt;
		for(int j = 1;j < cycle.size();j++) calc(cycle[j],-1);
		int tmp = a[cycle[0]];
		a[cycle[0]] = -1;
		calc(cycle[0],-1);
		if(tmp == a[cycle[0]]) flag = true;
		for(int j = 0;j < n;j++) a[j] = -1;
	}
	if(flag) cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
	return 0;
}
```