Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

4
2 3 4 1

Sample Output 1

POSSIBLE

The assignment is possible: {a_i} = {0, 1, 0, 1} or {a_i} = {1, 0, 1, 0}.

Sample Input 2

3
2 3 1

Sample Output 2

IMPOSSIBLE

Sample Input 3

4
2 3 1 1

Sample Output 3

POSSIBLE

The assignment is possible: {a_i} = {2, 0, 1, 0}.

Sample Input 4

6
4 5 6 5 6 4

Sample Output 4

IMPOSSIBLE

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int n;
vector<int> out[200010];
int in[200010];
int cycle[200010];
int grundy[200010];
int st = -1;
const int CYCLE = 42;

int dfs(int x, int p) {
	cycle[x] = 1;
	for(int i : out[x]) {
		if(i != p) {
			if(cycle[i] == 1) {
				st = i;
				cycle[x] = CYCLE;
				return CYCLE;
			}
			else if(cycle[i] == 0) {
				if(dfs(i, x) == CYCLE) {
					cycle[x] = CYCLE;
					if(x == st) return 0;
					else return CYCLE;
				}
			}
		}
	}
	cycle[x] = 2;
	return 0;
}

int getgrundy(int x) {
	set<int> s;
	for(int i : out[x])
		if(cycle[i] != CYCLE)
			s.insert(getgrundy(i));
	int ans = 0;
	auto it = s.begin();
	while(it != s.end()) {
		if(*it == ans) {
			++ans;
			++it;
		}
		else return grundy[x] = ans;
	}
	return grundy[x] = ans;
}

bool chk(int x, int val, int res) {
	if(x == st) return res == val;
	int nxt = -1;
	for(int i : out[x]) {
		if(cycle[i] == CYCLE) {
			nxt = i;
		}
	}
	if(nxt == -1) return false;
	return chk(nxt, grundy[nxt] + (grundy[nxt] == val), res);
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	memset(cycle, 0, sizeof cycle);
	for(int i = 0; i < n; ++i){
		int x;
		cin >> x;
		--x;
		out[x].push_back(i);
		in[i] = x;
		if(x < i and in[x] == i) {
			st = i;
			cycle[x] = cycle[i] = CYCLE;
		}
	}
	for(int i = 0; st == -1 and i < n; ++i) if(cycle[i] == 0) dfs(i, -1);
	for(int i = 0; i < n; ++i) {
		if(cycle[i] == CYCLE) getgrundy(i);
	}
	int nxt;
	for(int i : out[st]) {
		if(cycle[i] == CYCLE) {
			nxt = i;
		}
	}
	if(st == nxt or chk(nxt, grundy[nxt] + (grundy[st] == grundy[nxt]), grundy[st])) cout << "POSSIBLE" << endl;
	else if (chk(nxt, grundy[nxt] + ((grundy[st] + 1) == grundy[nxt]), grundy[st] + 1)) cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;

}