#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int gi()
{
	char c;int num=0,flg=1;
	while((c=getchar())<'0'||c>'9')if(c=='-')flg=-1;
	while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();}
	return num*flg;
}
#define N 200005
int fir[N],to[2*N],nxt[2*N],cnt;
void adde(int a,int b)
{
	to[++cnt]=b;nxt[cnt]=fir[a];fir[a]=cnt;
	to[++cnt]=a;nxt[cnt]=fir[b];fir[b]=cnt;
}
int fa[N],cir[N],clen,dep[N];bool onc[N];
void findcir(int u,int ff)
{
	fa[u]=ff;
	dep[u]=dep[ff]+1;
	for(int v,p=fir[u];p;p=nxt[p]){
		if((v=to[p])!=ff){
			if(!dep[v])findcir(v,u);
			else if(dep[v]<dep[u]){
				for(int t=u;;t=fa[t]){
					onc[t]=1;cir[++clen]=t;
					if(t==v)break;
				}
			}
		}
	}
}
int f[N];bool vis[N];
void dfs(int u,int ff)
{
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])dfs(v,u);
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])vis[f[v]]=1;//  KKK
	while(vis[f[u]])f[u]++;
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])vis[f[v]]=0;
}
int main()
{
	int n=gi(),i;
	for(i=1;i<=n;i++)adde(gi(),i);
	findcir(1,0);
	for(i=1;i<=clen;i++)dfs(cir[i],0);
	bool flg=0;
	for(i=2;i<=clen;i++)
		if(f[cir[i]]!=f[cir[1]]){flg=1;break;}
	if(!flg&&(clen&1))printf("IMPOSSIBLE\n");
	else printf("POSSIBLE\n");
}
### 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int gi()
{
	char c;int num=0,flg=1;
	while((c=getchar())<'0'||c>'9')if(c=='-')flg=-1;
	while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();}
	return num*flg;
}
#define N 200005
int fir[N],to[2*N],nxt[2*N],cnt;
void adde(int a,int b)
{
	to[++cnt]=b;nxt[cnt]=fir[a];fir[a]=cnt;
	to[++cnt]=a;nxt[cnt]=fir[b];fir[b]=cnt;
}
int fa[N],cir[N],clen,dep[N];bool onc[N];
void findcir(int u,int ff)
{
	fa[u]=ff;
	dep[u]=dep[ff]+1;
	for(int v,p=fir[u];p;p=nxt[p]){
		if((v=to[p])!=ff){
			if(!dep[v])findcir(v,u);
			else if(dep[v]<dep[u]){
				for(int t=u;;t=fa[t]){
					onc[t]=1;cir[++clen]=t;
					if(t==v)break;
				}
			}
		}
	}
}
int f[N];bool vis[N];
void dfs(int u,int ff)
{
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])dfs(v,u);
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])vis[f[v]]=1;//  KKK
	while(vis[f[u]])f[u]++;
	for(int v,p=fir[u];p;p=nxt[p])if((v=to[p])!=ff&&!onc[v])vis[f[v]]=0;
}
int main()
{
	int n=gi(),i;
	for(i=1;i<=n;i++)adde(gi(),i);
	findcir(1,0);
	for(i=1;i<=clen;i++)dfs(cir[i],0);
	bool flg=0;
	for(i=2;i<=clen;i++)
		if(f[cir[i]]!=f[cir[1]]){flg=1;break;}
	if(!flg&&(clen&1))printf("IMPOSSIBLE\n");
	else printf("POSSIBLE\n");
}
```