#include <stdio.h>
#include <vector>
#include <queue>
#include <set>
using namespace std;

set<int> S[200200],T[200200];
int N,G[200200],D[200200],M[200200];

int main()
{
	scanf ("%d",&N);
	for (int i=1;i<=N;i++){
		scanf ("%d",&G[i]);
		D[G[i]]++;
	}

	queue<int> Q;
	for (int i=1;i<=N;i++) if (D[i] == 0){
		Q.push(i);
	}

	while (!Q.empty()){
		int x = Q.front(); Q.pop();
		for (auto &m : S[x]){
			if (m == M[x]) M[x]++;
			else break;
		}
		S[G[x]].insert(M[x]);
		if (--D[G[x]] == 0) Q.push(G[x]);
	}

	for (int x=1;x<=N;x++) if (D[x] != 0){
		for (auto &m : S[x]){
			if (m == M[x]) M[x]++;
			else break;
		}

		for (int i=1;i<=N;i++) if (D[i] != 2) T[i] = S[i];
		int t = x;
		while (G[t] != x){
			T[G[t]].insert(M[t]);
			t = G[t];
			for (auto &m : T[t]){
				if (m == M[t]) M[t]++;
				else break;
			}
		}

		if (M[x] != M[t]){
			puts("POSSIBLE");
			return 0;
		}

		x = t;
		for (int i=1;i<=N;i++) if (D[i] != 2) T[i] = S[i];
		int o = M[t];
		while (true){
			T[G[t]].insert(M[t]);
			t = G[t];
			M[t] = 0;
			for (auto &m : T[t]){
				if (m == M[t]) M[t]++;
				else break;
			}
			if (t == x) break;
		}

		if (o == M[t]){
			puts("POSSIBLE");
			return 0;
		}
		break;
	}
	puts("IMPOSSIBLE");
	return 0;
}
### 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 <stdio.h>
#include <vector>
#include <queue>
#include <set>
using namespace std;

set<int> S[200200],T[200200];
int N,G[200200],D[200200],M[200200];

int main()
{
	scanf ("%d",&N);
	for (int i=1;i<=N;i++){
		scanf ("%d",&G[i]);
		D[G[i]]++;
	}

	queue<int> Q;
	for (int i=1;i<=N;i++) if (D[i] == 0){
		Q.push(i);
	}

	while (!Q.empty()){
		int x = Q.front(); Q.pop();
		for (auto &m : S[x]){
			if (m == M[x]) M[x]++;
			else break;
		}
		S[G[x]].insert(M[x]);
		if (--D[G[x]] == 0) Q.push(G[x]);
	}

	for (int x=1;x<=N;x++) if (D[x] != 0){
		for (auto &m : S[x]){
			if (m == M[x]) M[x]++;
			else break;
		}

		for (int i=1;i<=N;i++) if (D[i] != 2) T[i] = S[i];
		int t = x;
		while (G[t] != x){
			T[G[t]].insert(M[t]);
			t = G[t];
			for (auto &m : T[t]){
				if (m == M[t]) M[t]++;
				else break;
			}
		}

		if (M[x] != M[t]){
			puts("POSSIBLE");
			return 0;
		}

		x = t;
		for (int i=1;i<=N;i++) if (D[i] != 2) T[i] = S[i];
		int o = M[t];
		while (true){
			T[G[t]].insert(M[t]);
			t = G[t];
			M[t] = 0;
			for (auto &m : T[t]){
				if (m == M[t]) M[t]++;
				else break;
			}
			if (t == x) break;
		}

		if (o == M[t]){
			puts("POSSIBLE");
			return 0;
		}
		break;
	}
	puts("IMPOSSIBLE");
	return 0;
}
```