#include<bits/stdc++.h>

using namespace std;

#define N 300000

int n,a[N],d[N],sg[N];
bool vis[N];
vector<int> vec[N];
queue<int> q;

void solve(int x){
	d[x]=0;
	for (int i:vec[x]) vis[i]=1;
	for (int i=0;vis[i];++i) sg[x]=i+1;
	for (int i:vec[x]) vis[i]=0;
	int y=a[x],z=x;
	for (;d[y];y=a[y],z=a[z]){
		for (int i:vec[y]) vis[i]=1;
		vis[sg[z]]=1;
		for (int i=0;vis[i];++i) sg[y]=i+1;
		for (int i:vec[y]) vis[i]=0;
		vis[sg[z]]=0;
	}
	if (sg[x]!=sg[z]) return;
	int t=sg[x]; sg[x]=t+1;
	for (int i:vec[x]) vis[i]=1;
	for (int i=t+1;vis[i];++i) sg[x]=i+1;
	for (int i:vec[x]) vis[i]=0;
	y=a[x],z=x;
	for (;d[y];y=a[y],z=a[z]){
		for (int i:vec[y]) vis[i]=1;
		vis[sg[z]]=1;
		sg[y]=0;
		for (int i=0;vis[i];++i) sg[y]=i+1;
		for (int i:vec[y]) vis[i]=0;
		vis[sg[z]]=0;
	}
	if (sg[z]==t) return;
	puts("IMPOSSIBLE");
	exit(0);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;++i){
		scanf("%d",a+i);
		++d[a[i]];
	}
	for (int i=1;i<=n;++i)
		if (!d[i]) q.push(i);
	while (!q.empty()){
		int x=q.front(); q.pop();
		for (int i:vec[x]) vis[i]=1;
		for (int i=0;vis[i];++i) sg[x]=i+1;
		for (int i:vec[x]) vis[i]=0;
		vec[a[x]].push_back(sg[x]);
		if (!(--d[a[x]])) q.push(a[x]);
	}
	for (int i=1;i<=n;++i)
		if (d[i]){
			solve(i);
			break;
		}
	puts("POSSIBLE");
	
	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;

#define N 300000

int n,a[N],d[N],sg[N];
bool vis[N];
vector<int> vec[N];
queue<int> q;

void solve(int x){
	d[x]=0;
	for (int i:vec[x]) vis[i]=1;
	for (int i=0;vis[i];++i) sg[x]=i+1;
	for (int i:vec[x]) vis[i]=0;
	int y=a[x],z=x;
	for (;d[y];y=a[y],z=a[z]){
		for (int i:vec[y]) vis[i]=1;
		vis[sg[z]]=1;
		for (int i=0;vis[i];++i) sg[y]=i+1;
		for (int i:vec[y]) vis[i]=0;
		vis[sg[z]]=0;
	}
	if (sg[x]!=sg[z]) return;
	int t=sg[x]; sg[x]=t+1;
	for (int i:vec[x]) vis[i]=1;
	for (int i=t+1;vis[i];++i) sg[x]=i+1;
	for (int i:vec[x]) vis[i]=0;
	y=a[x],z=x;
	for (;d[y];y=a[y],z=a[z]){
		for (int i:vec[y]) vis[i]=1;
		vis[sg[z]]=1;
		sg[y]=0;
		for (int i=0;vis[i];++i) sg[y]=i+1;
		for (int i:vec[y]) vis[i]=0;
		vis[sg[z]]=0;
	}
	if (sg[z]==t) return;
	puts("IMPOSSIBLE");
	exit(0);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;++i){
		scanf("%d",a+i);
		++d[a[i]];
	}
	for (int i=1;i<=n;++i)
		if (!d[i]) q.push(i);
	while (!q.empty()){
		int x=q.front(); q.pop();
		for (int i:vec[x]) vis[i]=1;
		for (int i=0;vis[i];++i) sg[x]=i+1;
		for (int i:vec[x]) vis[i]=0;
		vec[a[x]].push_back(sg[x]);
		if (!(--d[a[x]])) q.push(a[x]);
	}
	for (int i=1;i<=n;++i)
		if (d[i]){
			solve(i);
			break;
		}
	puts("POSSIBLE");
	
	return 0;
}
			
```