#include<bits/stdc++.h>
using namespace std;
const int N=200006;
int n,t,p=0,cnt=0,maxn=-1,minn=1e9,f[N],v[N],s[N],g[N],fa[N],head[N];
struct edge{int nxt,to;}e[N<<1];
inline void add(int u,int v){e[++cnt].nxt=head[u],e[cnt].to=v,head[u]=cnt;} 
inline int read(){
  int T=0,F=1; char ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') F=-1; ch=getchar();}
  while(ch>='0'&&ch<='9') T=(T<<3)+(T<<1)+(ch-48),ch=getchar();
  return F*T; 
} 
void dfs(int x){
	f[x]=0;
    for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) dfs(e[i].to);
	for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) v[f[e[i].to]]=1;	
    while(v[f[x]]) ++f[x];
	for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) v[f[e[i].to]]=0;		
}
int main(){
	n=read(),t=1;
	for(int i=1;i<=n;++i) fa[i]=read(),add(fa[i],i);
	while(!s[t]) s[t]=1,t=fa[t];
	while(!g[t]) g[t]=1,t=fa[t];
	for(int i=1;i<=n;++i) if(g[i]) ++p,dfs(i),maxn=max(maxn,f[i]),minn=min(minn,f[i]);
	if(maxn==minn&&(p&1)) printf("IMPOSSIBLE\n");
	else printf("POSSIBLE"); 
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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=200006;
int n,t,p=0,cnt=0,maxn=-1,minn=1e9,f[N],v[N],s[N],g[N],fa[N],head[N];
struct edge{int nxt,to;}e[N<<1];
inline void add(int u,int v){e[++cnt].nxt=head[u],e[cnt].to=v,head[u]=cnt;} 
inline int read(){
  int T=0,F=1; char ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') F=-1; ch=getchar();}
  while(ch>='0'&&ch<='9') T=(T<<3)+(T<<1)+(ch-48),ch=getchar();
  return F*T; 
} 
void dfs(int x){
	f[x]=0;
    for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) dfs(e[i].to);
	for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) v[f[e[i].to]]=1;	
    while(v[f[x]]) ++f[x];
	for(int i=head[x];i;i=e[i].nxt) if(!g[e[i].to]) v[f[e[i].to]]=0;		
}
int main(){
	n=read(),t=1;
	for(int i=1;i<=n;++i) fa[i]=read(),add(fa[i],i);
	while(!s[t]) s[t]=1,t=fa[t];
	while(!g[t]) g[t]=1,t=fa[t];
	for(int i=1;i<=n;++i) if(g[i]) ++p,dfs(i),maxn=max(maxn,f[i]),minn=min(minn,f[i]);
	if(maxn==minn&&(p&1)) printf("IMPOSSIBLE\n");
	else printf("POSSIBLE"); 
	return 0;
}
```