#include <bits/stdc++.h>

using namespace std;

const int INF = 1e9;

vector< vector<int> > g;

bool dfs(int cur, vector<int>& used, vector<int>& cycle) {
	if (used[cur] == 1) {
		cycle[cur] = true;
		return true;
	}
	used[cur] = 1;
	for (int nex : g[cur]) {
		if (dfs(nex, used, cycle)) {
			cycle[cur] = true;
			return true;
		}
	}
	used[cur] = 2;
	return false;
}

int solve(int cur, vector<int>& cycle, vector<int>& grundy) {
	if (grundy[cur] != INF) return grundy[cur];
	set<int> s;
	for (int nex : g[cur]) {
		if (cycle[nex]) continue;
		int x = solve(nex, cycle, grundy);
		s.insert(x);
	}
	int res = 0;
	while (s.find(res) != s.end()) res++;
	return grundy[cur] = res;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	g.resize(n);
	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		p--;
		g[p].push_back(i);
	}
	vector<int> used(n, 0), cycle(n, false);
	for (int i = 0; i < n; i++) {
		if (used[i]) continue;
		dfs(i, used, cycle);
	}
	int x = -1, cnt = 0;
	bool ans = true;
	vector<int> grundy(n, INF);
	for (int i = 0; i < n; i++) {
		if (cycle[i]) {
			cnt++;
			int y = solve(i, cycle, grundy);
			if (x == -1) {
				x = y;
				ans = false;
			} else if (x != y) {
				ans = true;
				break;
			}
		}
	}
	if (!ans) ans = (cnt % 2 == 0);
	cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << endl;
	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 <bits/stdc++.h>

using namespace std;

const int INF = 1e9;

vector< vector<int> > g;

bool dfs(int cur, vector<int>& used, vector<int>& cycle) {
	if (used[cur] == 1) {
		cycle[cur] = true;
		return true;
	}
	used[cur] = 1;
	for (int nex : g[cur]) {
		if (dfs(nex, used, cycle)) {
			cycle[cur] = true;
			return true;
		}
	}
	used[cur] = 2;
	return false;
}

int solve(int cur, vector<int>& cycle, vector<int>& grundy) {
	if (grundy[cur] != INF) return grundy[cur];
	set<int> s;
	for (int nex : g[cur]) {
		if (cycle[nex]) continue;
		int x = solve(nex, cycle, grundy);
		s.insert(x);
	}
	int res = 0;
	while (s.find(res) != s.end()) res++;
	return grundy[cur] = res;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	g.resize(n);
	for (int i = 0; i < n; i++) {
		int p;
		cin >> p;
		p--;
		g[p].push_back(i);
	}
	vector<int> used(n, 0), cycle(n, false);
	for (int i = 0; i < n; i++) {
		if (used[i]) continue;
		dfs(i, used, cycle);
	}
	int x = -1, cnt = 0;
	bool ans = true;
	vector<int> grundy(n, INF);
	for (int i = 0; i < n; i++) {
		if (cycle[i]) {
			cnt++;
			int y = solve(i, cycle, grundy);
			if (x == -1) {
				x = y;
				ans = false;
			} else if (x != y) {
				ans = true;
				break;
			}
		}
	}
	if (!ans) ans = (cnt % 2 == 0);
	cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << endl;
	return 0;
}
```