#include <bits/stdc++.h>
using namespace std;
const int maxn = 200200, mx = 7;
vector<int> cycle;
int p[maxn], out[maxn], val[maxn], is_out[maxn];
set<int> has[maxn];
int main(){
	int n; cin >> n;
	for(int i = 0; i < n; i++){
		cin >> p[i]; p[i]--;
		out[p[i]]++;
	}
	set<pair<int,int> > outs;
	for(int i = 0; i < n; i++){
		outs.insert({out[i], i});
	}
	while(!outs.empty()){
		if(outs.begin()->first) break;
		int ot, v; tie(ot, v) = *outs.begin(); outs.erase(outs.begin());
		is_out[v] = true;
		while(has[v].count(val[v])) val[v]++;
		has[p[v]].insert(val[v]);
		outs.erase({out[p[v]], p[v]});
		out[p[v]]--;
		outs.insert({out[p[v]], p[v]});
	}
	for(int i = 0; i < n; i++){
		if(!is_out[i]){
			int u = i;
			do {
				cycle.push_back(u);
				u = p[u];
			} while(u != i);
			vector<int> cand;
			for(int j = 0; j < maxn && cand.size() < 2; j++) if(!has[cycle[0]].count(j)) cand.push_back(j);
			auto check = [&](int opt){
				int len = cycle.size();
				int lst = opt;
				for(int j = 0; j < len; j++){
					int v = 0, who = cycle[(j + 1) % len];
					while(has[who].count(v) || v == lst) v++;
					lst = v;
				}
				return lst == opt;
			};
			if(check(cand[0]) || check(cand[1])) cout << "POSSIBLE" << endl;
			else cout << "IMPOSSIBLE" << endl;
			return 0;
		}
	}
	return 0;
}
	


### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;
const int maxn = 200200, mx = 7;
vector<int> cycle;
int p[maxn], out[maxn], val[maxn], is_out[maxn];
set<int> has[maxn];
int main(){
	int n; cin >> n;
	for(int i = 0; i < n; i++){
		cin >> p[i]; p[i]--;
		out[p[i]]++;
	}
	set<pair<int,int> > outs;
	for(int i = 0; i < n; i++){
		outs.insert({out[i], i});
	}
	while(!outs.empty()){
		if(outs.begin()->first) break;
		int ot, v; tie(ot, v) = *outs.begin(); outs.erase(outs.begin());
		is_out[v] = true;
		while(has[v].count(val[v])) val[v]++;
		has[p[v]].insert(val[v]);
		outs.erase({out[p[v]], p[v]});
		out[p[v]]--;
		outs.insert({out[p[v]], p[v]});
	}
	for(int i = 0; i < n; i++){
		if(!is_out[i]){
			int u = i;
			do {
				cycle.push_back(u);
				u = p[u];
			} while(u != i);
			vector<int> cand;
			for(int j = 0; j < maxn && cand.size() < 2; j++) if(!has[cycle[0]].count(j)) cand.push_back(j);
			auto check = [&](int opt){
				int len = cycle.size();
				int lst = opt;
				for(int j = 0; j < len; j++){
					int v = 0, who = cycle[(j + 1) % len];
					while(has[who].count(v) || v == lst) v++;
					lst = v;
				}
				return lst == opt;
			};
			if(check(cand[0]) || check(cand[1])) cout << "POSSIBLE" << endl;
			else cout << "IMPOSSIBLE" << endl;
			return 0;
		}
	}
	return 0;
}
	


```