#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define For(i,u) for(int i=hed[u],v=e[i].to;i;i=e[i].nex,v=e[i].to)
using namespace std;
const int maxn=2e5+10;
int n,p[maxn],hed[maxn],cnt,f[maxn],sg[maxn],a[maxn],vis[maxn],ed;
struct edge{
	int to,nex;
	edge(int _=0,int __=0):to(_),nex(__){}
}e[maxn];
inline void add(int u,int v){e[++ed]=edge(v,hed[u]);hed[u]=ed;}
void get_cir(){
	int u=1;
	while(!vis[u]) vis[u]=1,u=p[u];
	memset(vis,0,sizeof(vis));
	while(!vis[u]) vis[u]=1,f[++cnt]=u,u=p[u];
}
void dfs(int u){
	For(i,u) if(!vis[v]) dfs(v);
	For(i,u) if(!vis[v]) a[sg[v]]=u;
	for(;a[sg[u]]==u;++sg[u]);
}
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",&p[i]),add(p[i],i);
	get_cir();int mx=-1e9,mn=1e9;
	rep(i,1,cnt)
		dfs(f[i]),mx=max(sg[f[i]],mx),mn=min(sg[f[i]],mn);
	if(mx==mn&&(cnt&1)) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define For(i,u) for(int i=hed[u],v=e[i].to;i;i=e[i].nex,v=e[i].to)
using namespace std;
const int maxn=2e5+10;
int n,p[maxn],hed[maxn],cnt,f[maxn],sg[maxn],a[maxn],vis[maxn],ed;
struct edge{
	int to,nex;
	edge(int _=0,int __=0):to(_),nex(__){}
}e[maxn];
inline void add(int u,int v){e[++ed]=edge(v,hed[u]);hed[u]=ed;}
void get_cir(){
	int u=1;
	while(!vis[u]) vis[u]=1,u=p[u];
	memset(vis,0,sizeof(vis));
	while(!vis[u]) vis[u]=1,f[++cnt]=u,u=p[u];
}
void dfs(int u){
	For(i,u) if(!vis[v]) dfs(v);
	For(i,u) if(!vis[v]) a[sg[v]]=u;
	for(;a[sg[u]]==u;++sg[u]);
}
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",&p[i]),add(p[i],i);
	get_cir();int mx=-1e9,mn=1e9;
	rep(i,1,cnt)
		dfs(f[i]),mx=max(sg[f[i]],mx),mn=min(sg[f[i]],mn);
	if(mx==mn&&(cnt&1)) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}
```