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

const int MAXN = 200001;

int n;
int p[MAXN];
vector<int> nxt[MAXN];
vector<int> cycle;

int vis[MAXN];
void find_cycle(int v) {
	vis[v] = 1;
	for(int w : nxt[v]) {
		if(vis[w] == 1) {
			cycle.push_back(w);
		}	
		if(!vis[w]) find_cycle(w);
	}
	vis[v] = 2;
}

bool marked[MAXN];
int mark[MAXN];
bool in_cycle[MAXN];

void build(int v) {
	set<int> used;
	for(int w : nxt[v]) {
		if(in_cycle[w]) continue;
		build(w);
		used.insert(mark[w]);
	}		
	for(int x = 0;; x++) {
		if(used.find(x) == used.end()) {
			mark[v] = x;
			break;
		}
	}
	marked[v] = used.size() > 0;
}



int main() {
	
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> p[i];
		nxt[p[i]].push_back(i);
	}
	
	for(int v = 1; v <= n; v++) {
		if(!vis[v]) find_cycle(v);
	}
	
	in_cycle[cycle[0]] = 1;
	
	for(int i = 0; i < (int)cycle.size(); i++) {
		int pv = p[cycle[i]];
		if(pv != cycle[0]) {
			cycle.push_back(pv);
			in_cycle[pv] = 1;
		}
	}
	
	for(int v : cycle) {
		build(v);
	}
	
	int m = cycle.size();
	for(int i = 0; i < m; i++) cycle.push_back(cycle[i]);

	int pv = 0;
	for(int v : cycle) {
		if(marked[v]) pv = v;
	}
		
	if(!pv && (n&1)) {
		cout << "IMPOSSIBLE\n";
		return 0;
	}
	
	int v = p[pv];
	for(int i = 0; i < 3*m; i++) {
		int pmark = mark[v];
		set<int> used;
		for(int w : nxt[v]) used.insert(mark[w]);
		for(int x = 0;; x++) {
			if(used.find(x) == used.end()) {
				mark[v] = x;
				break;
			}
		}
			
		if(i > 2*m && mark[v] != pmark) {
			cout << "IMPOSSIBLE\n";
			return 0;
		}
		v = p[v];
	}
	
	
	cout << "POSSIBLE\n";
	
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 200001;

int n;
int p[MAXN];
vector<int> nxt[MAXN];
vector<int> cycle;

int vis[MAXN];
void find_cycle(int v) {
	vis[v] = 1;
	for(int w : nxt[v]) {
		if(vis[w] == 1) {
			cycle.push_back(w);
		}	
		if(!vis[w]) find_cycle(w);
	}
	vis[v] = 2;
}

bool marked[MAXN];
int mark[MAXN];
bool in_cycle[MAXN];

void build(int v) {
	set<int> used;
	for(int w : nxt[v]) {
		if(in_cycle[w]) continue;
		build(w);
		used.insert(mark[w]);
	}		
	for(int x = 0;; x++) {
		if(used.find(x) == used.end()) {
			mark[v] = x;
			break;
		}
	}
	marked[v] = used.size() > 0;
}



int main() {
	
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> p[i];
		nxt[p[i]].push_back(i);
	}
	
	for(int v = 1; v <= n; v++) {
		if(!vis[v]) find_cycle(v);
	}
	
	in_cycle[cycle[0]] = 1;
	
	for(int i = 0; i < (int)cycle.size(); i++) {
		int pv = p[cycle[i]];
		if(pv != cycle[0]) {
			cycle.push_back(pv);
			in_cycle[pv] = 1;
		}
	}
	
	for(int v : cycle) {
		build(v);
	}
	
	int m = cycle.size();
	for(int i = 0; i < m; i++) cycle.push_back(cycle[i]);

	int pv = 0;
	for(int v : cycle) {
		if(marked[v]) pv = v;
	}
		
	if(!pv && (n&1)) {
		cout << "IMPOSSIBLE\n";
		return 0;
	}
	
	int v = p[pv];
	for(int i = 0; i < 3*m; i++) {
		int pmark = mark[v];
		set<int> used;
		for(int w : nxt[v]) used.insert(mark[w]);
		for(int x = 0;; x++) {
			if(used.find(x) == used.end()) {
				mark[v] = x;
				break;
			}
		}
			
		if(i > 2*m && mark[v] != pmark) {
			cout << "IMPOSSIBLE\n";
			return 0;
		}
		v = p[v];
	}
	
	
	cout << "POSSIBLE\n";
	
	return 0;
}
```