#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
vector<int> edges[210000];
vector<int> redges[210000];
int n;
int p[210000];
int cyc[210000];
int vis[210000];
int t;
void dfs(int a, int st){
	if(vis[a]){
		if(a == st){
			t = st;
		}
		return;
	}
	vis[a] = 1;
	for(int j = 0; j < redges[a].size(); j++){
		dfs(redges[a][j], st);
	}
}

int gr[210000];
void dfs2(int a){
	if(gr[a] != -1) return;
	gr[a] = 0;
	map<int,int> d;
	for(int j = 0; j < redges[a].size(); j++){
		if(cyc[redges[a][j]]) continue;
		dfs2(redges[a][j]);
		d[gr[redges[a][j]]]++;
	}
	while(d[gr[a]]){
		gr[a]++;
	}
}

int ok = 0;

int time0;
void tryval(int a, int g, int st, int init){
	//cout << a << " " << g << endl;
	if(a == st && time0){
		if(g == init) ok = 1;
		return;
	}
	time0 = 1;
	map<int,int> d;
	int pa = p[a];
	for(int j = 0; j < redges[pa].size(); j++){
		if(cyc[redges[pa][j]]) continue;
		d[gr[redges[pa][j]]]++;
	}
	d[g]++;
	int b = 0;
	while(d[b]) b++;
	tryval(p[a], b, st, init);
}

int main(){
	cin >> n;
	for(int i = 0; i < n; i++){
		vis[i] = 0;
		gr[i] = -1;
		cyc[i] = 0;
		int a;
		cin >> a;
		a--;
		p[i] = a;
		edges[i].push_back(a);
		redges[a].push_back(i);
	}

	for(int i = 0; i < n; i++){
		if(!vis[i]) dfs(i, i);
	}
	int s = t;
	int ans = 0;
	while(1){
		cyc[s] = 1;
		s = p[s];
		ans++;
		if(s == t){
			break;
		}
	}

	//cout << ans << endl;
	for(int i = 0; i < n; i++){
		if(cyc[i]) dfs2(i);
	}

	time0 = 0;
	tryval(p[t], gr[p[t]], p[t], gr[p[t]]);
	//cout << "here" << p[t] << " " << gr[p[t]] << endl;
	time0 = 0;
	tryval(t, gr[p[t]], t, gr[p[t]]);
	//cout << "here" << t << " " << gr[p[t]] << endl;

	cout << (ok ? "POSSIBLE" : "IMPOSSIBLE") << endl;
}

### 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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
vector<int> edges[210000];
vector<int> redges[210000];
int n;
int p[210000];
int cyc[210000];
int vis[210000];
int t;
void dfs(int a, int st){
	if(vis[a]){
		if(a == st){
			t = st;
		}
		return;
	}
	vis[a] = 1;
	for(int j = 0; j < redges[a].size(); j++){
		dfs(redges[a][j], st);
	}
}

int gr[210000];
void dfs2(int a){
	if(gr[a] != -1) return;
	gr[a] = 0;
	map<int,int> d;
	for(int j = 0; j < redges[a].size(); j++){
		if(cyc[redges[a][j]]) continue;
		dfs2(redges[a][j]);
		d[gr[redges[a][j]]]++;
	}
	while(d[gr[a]]){
		gr[a]++;
	}
}

int ok = 0;

int time0;
void tryval(int a, int g, int st, int init){
	//cout << a << " " << g << endl;
	if(a == st && time0){
		if(g == init) ok = 1;
		return;
	}
	time0 = 1;
	map<int,int> d;
	int pa = p[a];
	for(int j = 0; j < redges[pa].size(); j++){
		if(cyc[redges[pa][j]]) continue;
		d[gr[redges[pa][j]]]++;
	}
	d[g]++;
	int b = 0;
	while(d[b]) b++;
	tryval(p[a], b, st, init);
}

int main(){
	cin >> n;
	for(int i = 0; i < n; i++){
		vis[i] = 0;
		gr[i] = -1;
		cyc[i] = 0;
		int a;
		cin >> a;
		a--;
		p[i] = a;
		edges[i].push_back(a);
		redges[a].push_back(i);
	}

	for(int i = 0; i < n; i++){
		if(!vis[i]) dfs(i, i);
	}
	int s = t;
	int ans = 0;
	while(1){
		cyc[s] = 1;
		s = p[s];
		ans++;
		if(s == t){
			break;
		}
	}

	//cout << ans << endl;
	for(int i = 0; i < n; i++){
		if(cyc[i]) dfs2(i);
	}

	time0 = 0;
	tryval(p[t], gr[p[t]], p[t], gr[p[t]]);
	//cout << "here" << p[t] << " " << gr[p[t]] << endl;
	time0 = 0;
	tryval(t, gr[p[t]], t, gr[p[t]]);
	//cout << "here" << t << " " << gr[p[t]] << endl;

	cout << (ok ? "POSSIBLE" : "IMPOSSIBLE") << endl;
}

```