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

const int N=200100;
int clk,cnt,a[N],fa[N],n,vis[N],sg[N],tmp[N];
vector<int>g[N];

void dfs(int x){
	for(int i:g[x])if(!vis[i])dfs(i);
	++clk;
	for(int i:g[x])if(!vis[i])tmp[sg[i]]=clk;
	for(sg[x]=0;tmp[sg[x]]==clk;++sg[x]);
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>fa[i],g[fa[i]].push_back(i);
	int rt=1;
	for(int i=1;i<=n;++i)rt=fa[rt];
	for(int i=1;i<=n;++i)vis[rt]=1,rt=fa[rt];
	for(int i=1;i<=n;++i)
		if(vis[i]==1)dfs(i),a[++cnt]=sg[i];
	sort(a+1,a+cnt+1);
	cout<<(a[1]!=a[cnt]||(a[1]==a[cnt]&&cnt%2==0)?"POSSIBLE\n":"IMPOSSIBLE\n");
	return 0;
}

### Prompt

Your challenge is to write 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;

const int N=200100;
int clk,cnt,a[N],fa[N],n,vis[N],sg[N],tmp[N];
vector<int>g[N];

void dfs(int x){
	for(int i:g[x])if(!vis[i])dfs(i);
	++clk;
	for(int i:g[x])if(!vis[i])tmp[sg[i]]=clk;
	for(sg[x]=0;tmp[sg[x]]==clk;++sg[x]);
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>fa[i],g[fa[i]].push_back(i);
	int rt=1;
	for(int i=1;i<=n;++i)rt=fa[rt];
	for(int i=1;i<=n;++i)vis[rt]=1,rt=fa[rt];
	for(int i=1;i<=n;++i)
		if(vis[i]==1)dfs(i),a[++cnt]=sg[i];
	sort(a+1,a+cnt+1);
	cout<<(a[1]!=a[cnt]||(a[1]==a[cnt]&&cnt%2==0)?"POSSIBLE\n":"IMPOSSIBLE\n");
	return 0;
}

```