#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<functional>
using namespace std;
vector<int>x[220000], V; int n, p[220000], q[220000], r[220000], u[220000];
int dfs(int pos) {
	vector<int>R(x[pos].size(), 0);
	for (int i = 0; i < x[pos].size(); i++) {
		if (q[x[pos][i]] == 2)continue;
		int s = dfs(x[pos][i]);
		if (s < x[pos].size())R[s] = 1;
	}
	for (int i = 0; i < x[pos].size(); i++) { if (R[i] == 0)return i; }
	return x[pos].size();
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> p[i];
		x[p[i]].push_back(i);
	}
	int cx = 1;
	while (true) {
		if (q[cx] == 2)break;
		q[cx]++; if (q[cx] == 2)V.push_back(cx);
		cx = p[cx];
	}
	for (int i = 0; i < V.size(); i++)r[i] = dfs(V[i]);
	u[0] = r[0];
	for (int i = 1; i < V.size(); i++) {
		if (r[i] == u[i - 1]) { u[i] = r[i] + 1; }
		else { u[i] = r[i]; }
	}
	bool OK = false;
	if (u[0] != u[V.size() - 1]) { OK = true; }
	u[0] = r[0] + 1;
	for (int i = 1; i < V.size(); i++) {
		if (r[i] == u[i - 1]) { u[i] = r[i] + 1; }
		else { u[i] = r[i]; }
	}
	if (u[0] == u[V.size() - 1] + 1) { OK = true; }
	if (OK == true)cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
	return 0;
}
### Prompt

Your challenge is to write 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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<functional>
using namespace std;
vector<int>x[220000], V; int n, p[220000], q[220000], r[220000], u[220000];
int dfs(int pos) {
	vector<int>R(x[pos].size(), 0);
	for (int i = 0; i < x[pos].size(); i++) {
		if (q[x[pos][i]] == 2)continue;
		int s = dfs(x[pos][i]);
		if (s < x[pos].size())R[s] = 1;
	}
	for (int i = 0; i < x[pos].size(); i++) { if (R[i] == 0)return i; }
	return x[pos].size();
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> p[i];
		x[p[i]].push_back(i);
	}
	int cx = 1;
	while (true) {
		if (q[cx] == 2)break;
		q[cx]++; if (q[cx] == 2)V.push_back(cx);
		cx = p[cx];
	}
	for (int i = 0; i < V.size(); i++)r[i] = dfs(V[i]);
	u[0] = r[0];
	for (int i = 1; i < V.size(); i++) {
		if (r[i] == u[i - 1]) { u[i] = r[i] + 1; }
		else { u[i] = r[i]; }
	}
	bool OK = false;
	if (u[0] != u[V.size() - 1]) { OK = true; }
	u[0] = r[0] + 1;
	for (int i = 1; i < V.size(); i++) {
		if (r[i] == u[i - 1]) { u[i] = r[i] + 1; }
		else { u[i] = r[i]; }
	}
	if (u[0] == u[V.size() - 1] + 1) { OK = true; }
	if (OK == true)cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
	return 0;
}
```