#include<bits/stdc++.h>
using namespace std;
template<typename TP>inline void read(TP &tar)
{
	TP ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	tar=ret*f;
}
namespace RKK
{
const int N=200011;
struct sumireko{int to,ne;}e[N];int he[N],ecnt;
void addline(int f,int t){e[++ecnt].to=t;e[ecnt].ne=he[f],he[f]=ecnt;}
int n,fa[N];
int rnd[N],ra,onr[N];
bool vis[N];
void findrnd()
{
	int x=1;
	while(!vis[x]) vis[x]=1,x=fa[x];
	while(!onr[x]) onr[x]=1,rnd[++ra]=x,x=fa[x];
}
int sg[N],sgvis[N],ipa;
void dfs(int x)
{
	for(int i=he[x],t=e[i].to;i;i=e[i].ne,t=e[i].to)if(!onr[t])
		dfs(t);
	ipa++;
	for(int i=he[x],t=e[i].to;i;i=e[i].ne,t=e[i].to)if(!onr[t])
		sgvis[sg[t]]=ipa;
	for(int i=0;;i++)if(sgvis[i]!=ipa){sg[x]=i;break;}
}

int Iris()
{
	read(n);for(int i=1;i<=n;i++) read(fa[i]),addline(fa[i],i);
	findrnd();
	int ma=-1,mi=n;
	if(ra&1)
	{
		for(int i=1,x=rnd[i];i<=ra;i++,x=rnd[i])
			dfs(x),ma=max(ma,sg[x]),mi=min(mi,sg[x]);
		puts(ma==mi?"IMPOSSIBLE":"POSSIBLE");
	}else puts("POSSIBLE");
	return 0;
}
}
int main(){return RKK::Iris();}
### Prompt

Generate 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;
template<typename TP>inline void read(TP &tar)
{
	TP ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	tar=ret*f;
}
namespace RKK
{
const int N=200011;
struct sumireko{int to,ne;}e[N];int he[N],ecnt;
void addline(int f,int t){e[++ecnt].to=t;e[ecnt].ne=he[f],he[f]=ecnt;}
int n,fa[N];
int rnd[N],ra,onr[N];
bool vis[N];
void findrnd()
{
	int x=1;
	while(!vis[x]) vis[x]=1,x=fa[x];
	while(!onr[x]) onr[x]=1,rnd[++ra]=x,x=fa[x];
}
int sg[N],sgvis[N],ipa;
void dfs(int x)
{
	for(int i=he[x],t=e[i].to;i;i=e[i].ne,t=e[i].to)if(!onr[t])
		dfs(t);
	ipa++;
	for(int i=he[x],t=e[i].to;i;i=e[i].ne,t=e[i].to)if(!onr[t])
		sgvis[sg[t]]=ipa;
	for(int i=0;;i++)if(sgvis[i]!=ipa){sg[x]=i;break;}
}

int Iris()
{
	read(n);for(int i=1;i<=n;i++) read(fa[i]),addline(fa[i],i);
	findrnd();
	int ma=-1,mi=n;
	if(ra&1)
	{
		for(int i=1,x=rnd[i];i<=ra;i++,x=rnd[i])
			dfs(x),ma=max(ma,sg[x]),mi=min(mi,sg[x]);
		puts(ma==mi?"IMPOSSIBLE":"POSSIBLE");
	}else puts("POSSIBLE");
	return 0;
}
}
int main(){return RKK::Iris();}
```