#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<lint, lint> pi;
const int MAXN = 200005;

vector<int> gph[MAXN];
int n, a[MAXN], deg[MAXN];
int g1[MAXN], g2[MAXN];
int ans[MAXN];

int dfs(int x){
	set<int> s;
	for(auto &i : gph[x]){
		s.insert(dfs(i));
	}
	for(int i=0; ; i++){
		if(s.find(i) == s.end()) return i;
	}
}

bool solve(int x){
	int prv = x;
	for(int j=a[x]; ; j=a[j]){
		int tmp = (ans[prv] == g1[j] ? g2[j] : g1[j]);
		if(j == x){
			return tmp == ans[j];
		}
		ans[j] = tmp;
		prv = j;
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%d",&a[i]);
		gph[a[i]].push_back(i);
		deg[a[i]]++;
	}
	queue<int> que;
	for(int i=1; i<=n; i++) if(!deg[i]) que.push(i);
	while(!que.empty()){
		int x = que.front();
		que.pop();
		deg[a[x]]--;
		if(!deg[a[x]]) que.push(a[x]);
	}
	int piv = 0;
	for(int i=1; i<=n; i++){
		if(!deg[i]) continue;
		piv = i;
		set<int> s;
		for(auto &j : gph[i]){
			if(!deg[j]) s.insert(dfs(j));
		}
		int cnt = 0;
		for(int j=0; ; j++){
			if(s.find(j) == s.end()){
				if(cnt == 0) g1[i] = j;
				if(cnt == 1) g2[i] = j;
				cnt++;
			}
			if(cnt == 2) break;
		}
	}
	ans[piv] = g1[piv];
	if(solve(piv)){
		puts("POSSIBLE");
		return 0;
	}
	ans[piv] = g2[piv];
	if(solve(piv)){
		puts("POSSIBLE");
		return 0;
	}
	puts("IMPOSSIBLE");
}

### Prompt

Your task is to create a cpp solution 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;
typedef long long lint;
typedef pair<lint, lint> pi;
const int MAXN = 200005;

vector<int> gph[MAXN];
int n, a[MAXN], deg[MAXN];
int g1[MAXN], g2[MAXN];
int ans[MAXN];

int dfs(int x){
	set<int> s;
	for(auto &i : gph[x]){
		s.insert(dfs(i));
	}
	for(int i=0; ; i++){
		if(s.find(i) == s.end()) return i;
	}
}

bool solve(int x){
	int prv = x;
	for(int j=a[x]; ; j=a[j]){
		int tmp = (ans[prv] == g1[j] ? g2[j] : g1[j]);
		if(j == x){
			return tmp == ans[j];
		}
		ans[j] = tmp;
		prv = j;
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%d",&a[i]);
		gph[a[i]].push_back(i);
		deg[a[i]]++;
	}
	queue<int> que;
	for(int i=1; i<=n; i++) if(!deg[i]) que.push(i);
	while(!que.empty()){
		int x = que.front();
		que.pop();
		deg[a[x]]--;
		if(!deg[a[x]]) que.push(a[x]);
	}
	int piv = 0;
	for(int i=1; i<=n; i++){
		if(!deg[i]) continue;
		piv = i;
		set<int> s;
		for(auto &j : gph[i]){
			if(!deg[j]) s.insert(dfs(j));
		}
		int cnt = 0;
		for(int j=0; ; j++){
			if(s.find(j) == s.end()){
				if(cnt == 0) g1[i] = j;
				if(cnt == 1) g2[i] = j;
				cnt++;
			}
			if(cnt == 2) break;
		}
	}
	ans[piv] = g1[piv];
	if(solve(piv)){
		puts("POSSIBLE");
		return 0;
	}
	ans[piv] = g2[piv];
	if(solve(piv)){
		puts("POSSIBLE");
		return 0;
	}
	puts("IMPOSSIBLE");
}

```