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;
const int MAX_N=2e5+10;

int N,pos=-1;
vector<int> G[MAX_N];
bool seen[MAX_N],finished[MAX_N];
stack<int> hist;
set<int> cycle;
int par[MAX_N],grundy[MAX_N][2];

void dfs1(int v){
    seen[v]=true; hist.emplace(v);
    for (int u:G[v]){
        if (finished[u]) continue;
        if (seen[u]&&!finished[u]){
            pos=u; return;
        }
        dfs1(u);
        if (~pos) return;
    }
    finished[v]=true; hist.pop();
}

void dfs2(int v,int p){
    set<int> mex;
    for (int u:G[v]){
        if (u==p) continue;
        if (p<0&&cycle.count(u)) continue;
        dfs2(u,v);
        mex.emplace(grundy[u][0]);
    }
    if (~p){
        for (int i=0;;++i){
            if (!mex.count(i)){
                grundy[v][0]=i;
                break;
            }
        }
    } else {
        int ok=0;
        for (int i=0;;++i){
            if (!mex.count(i)){
                if (ok){grundy[v][1]=i; break;}
                else {grundy[v][0]=i; ok=1;}
            }
        }
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    for (int i=0;i<N;++i){
        cin >> par[i];
        G[--par[i]].emplace_back(i);
    }
    for (int i=0;i<N;++i){
        if (!seen[i]) dfs1(i);
        if (~pos) break;
    }
    while(!hist.empty()){
        int t=hist.top(); hist.pop();
        cycle.emplace(t);
        if (t==pos) break;
    }
    for (int v:cycle) dfs2(v,-1);
    int s=*cycle.begin();
    for (int i=0;i<2;++i){
        vector<int> res(N);
        int now=s; res[s]=grundy[s][i];
        while(par[now]!=s){
            int p=par[now];
            res[p]=(res[now]==grundy[p][0]?grundy[p][1]:grundy[p][0]);
            now=p;
        }
        if ((res[now]==grundy[s][0])==i){
            cout << "POSSIBLE" << '\n';
            return 0;
        }
    }
    cout << "IMPOSSIBLE" << '\n';
}