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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;

int n, in[MAXN], deg[MAXN], val[MAXN];
vector <int> mex[MAXN]; queue <int> q;

int calc(vector <int> &cal){
	int ret;
	for(ret = 0; ret < int(cal.size()) && cal[ret]; ++ ret) ;
	return ret;
}
bool check(int p, int v){
	int i; val[p] = v;
	for(i = p; in[i] != p; i = in[i]){
		if(int(mex[in[i]].size()) <= val[i])
			mex[in[i]].resize(val[i] + 1);
		++ mex[in[i]][val[i]];
		val[in[i]] = calc(mex[in[i]]);
	}
	if(int(mex[in[i]].size()) <= val[i])
		mex[in[i]].resize(val[i] + 1);
	++ mex[in[i]][val[i]];
	bool ret = calc(mex[in[i]]) == val[in[i]];
	for(i = p; in[i] != p; i = in[i])
		-- mex[in[i]][val[i]];
	-- mex[in[i]][val[i]];
	return ret;
}
int main(){
	int i;
	scanf("%d", &n);
	for(i = 1; i <= n; ++ i){
		scanf("%d", &in[i]);
		++ deg[in[i]];
	}
	memset(val, -1, sizeof(val));
	for(i = 1; i <= n; ++ i)
		if(!deg[i]) q.push(i);
	while(!q.empty()){
		i = q.front(); q.pop();
		val[i] = calc(mex[i]);
		if(int(mex[in[i]].size()) <= val[i])
			mex[in[i]].resize(val[i] + 1);
		++ mex[in[i]][val[i]];
		if(!(-- deg[in[i]])) q.push(in[i]);
	}
	for(i = 1; i <= n; ++ i)
		if(val[i] == -1) break;
	printf("%s\n", check(i, calc(mex[i])) || check(i, calc(mex[i]) + 1) ? "POSSIBLE" : "IMPOSSIBLE");
	return 0;
}