#include<bits/stdc++.h>
using namespace std;
const int N=200005;
int n,vis[N],p[N];
int f[N],g[N],dp[N],tong[N];
vector<int> v[N];
void dfs(int x){
	for (int i=0;i<v[x].size();i++) dfs(v[x][i]);
	for (int i=0;i<v[x].size();i++)
		tong[dp[v[x][i]]]++;
	for (;tong[dp[x]];dp[x]++);
	for (int i=0;i<v[x].size();i++)
		tong[dp[v[x][i]]]--;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&p[i]);
		v[p[i]].push_back(i);
	}
	for (int i=1;;i=p[i]){
		vis[i]++;
		if (vis[i]>2) break;
	}
	for (int i=1;i<=n;i++)
		if (vis[i]>=2){
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					dfs(v[i][j]);
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					tong[dp[v[i][j]]]++;
			f[i]=g[i]=-1;
			for (int j=0;;j++)
				if (!tong[j]){
					if (f[i]==-1) f[i]=j;
					else if (g[i]==-1) g[i]=j;
					else break;
				}
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					tong[dp[v[i][j]]]--;
		}
	for (int i=1;i<=n;i++)
		if (vis[i]>=2){
			dp[i]=f[i];
			for (int j=i;;j=p[j]){
				if (p[j]==i){
					if (dp[j]^dp[i])
						return puts("POSSIBLE"),0;
					break;
				}
				dp[p[j]]=(f[p[j]]==dp[j]?g[p[j]]:f[p[j]]);
			}
			dp[i]=g[i];
			for (int j=i;;j=p[j]){
				if (p[j]==i){
					if (dp[j]==f[i])
						return puts("POSSIBLE"),0;
					break;
				}
				dp[p[j]]=(f[p[j]]==dp[j]?g[p[j]]:f[p[j]]);
			}
			return puts("IMPOSSIBLE"),0;
		}
}
### Prompt

In Cpp, your task is to solve 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 N=200005;
int n,vis[N],p[N];
int f[N],g[N],dp[N],tong[N];
vector<int> v[N];
void dfs(int x){
	for (int i=0;i<v[x].size();i++) dfs(v[x][i]);
	for (int i=0;i<v[x].size();i++)
		tong[dp[v[x][i]]]++;
	for (;tong[dp[x]];dp[x]++);
	for (int i=0;i<v[x].size();i++)
		tong[dp[v[x][i]]]--;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&p[i]);
		v[p[i]].push_back(i);
	}
	for (int i=1;;i=p[i]){
		vis[i]++;
		if (vis[i]>2) break;
	}
	for (int i=1;i<=n;i++)
		if (vis[i]>=2){
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					dfs(v[i][j]);
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					tong[dp[v[i][j]]]++;
			f[i]=g[i]=-1;
			for (int j=0;;j++)
				if (!tong[j]){
					if (f[i]==-1) f[i]=j;
					else if (g[i]==-1) g[i]=j;
					else break;
				}
			for (int j=0;j<v[i].size();j++)
				if (vis[v[i][j]]<2)
					tong[dp[v[i][j]]]--;
		}
	for (int i=1;i<=n;i++)
		if (vis[i]>=2){
			dp[i]=f[i];
			for (int j=i;;j=p[j]){
				if (p[j]==i){
					if (dp[j]^dp[i])
						return puts("POSSIBLE"),0;
					break;
				}
				dp[p[j]]=(f[p[j]]==dp[j]?g[p[j]]:f[p[j]]);
			}
			dp[i]=g[i];
			for (int j=i;;j=p[j]){
				if (p[j]==i){
					if (dp[j]==f[i])
						return puts("POSSIBLE"),0;
					break;
				}
				dp[p[j]]=(f[p[j]]==dp[j]?g[p[j]]:f[p[j]]);
			}
			return puts("IMPOSSIBLE"),0;
		}
}
```