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

#define N 200010
#define inf 0x3f3f3f3f
int n;
int seg[N],used[N],fa[N],cir[N];
int first[N],cnt=0;
struct edge{
	int u,v,nxt;
}e[N<<1];

inline void add(int u,int v){e[++cnt].u=u;e[cnt].v=v;e[cnt].nxt=first[u];first[u]=cnt;}

inline void dfs(int u){
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[v]) dfs(v);
	}
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[v]) used[seg[v]]=1;//由外向树的叶子节点向环上点更新 
	}
	for(;used[seg[u]];seg[u]++);//到此说明u的子树已经被统计完了,要满足条件2,[0,u)要包含u所有子节点的权值 
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[i]) used[seg[v]]=0;//标记清0 
	}
}

int main(){
	scanf("%d",&n);
	for(int p,i=1;i<=n;i++){scanf("%d",&p);add(p,i);fa[i]=p;}
	int now=1;
	for(;!used[now];now=fa[now]) used[now]=1;
	for(;!cir[now];now=fa[now]) cir[now]=1;
	memset(used,0,sizeof(used));
	int num=0,maxi=-inf,mini=inf;
	for(int i=1;i<=n;i++)
		if(cir[i]){
			dfs(i);
			num++;
			maxi=max(maxi,seg[i]);
			mini=min(mini,seg[i]);
		}
	if(maxi==mini&&num%2) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}
### Prompt

Your task is to create 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;

#define N 200010
#define inf 0x3f3f3f3f
int n;
int seg[N],used[N],fa[N],cir[N];
int first[N],cnt=0;
struct edge{
	int u,v,nxt;
}e[N<<1];

inline void add(int u,int v){e[++cnt].u=u;e[cnt].v=v;e[cnt].nxt=first[u];first[u]=cnt;}

inline void dfs(int u){
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[v]) dfs(v);
	}
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[v]) used[seg[v]]=1;//由外向树的叶子节点向环上点更新 
	}
	for(;used[seg[u]];seg[u]++);//到此说明u的子树已经被统计完了,要满足条件2,[0,u)要包含u所有子节点的权值 
	for(int i=first[u];i;i=e[i].nxt){
		int v=e[i].v;
		if(!cir[i]) used[seg[v]]=0;//标记清0 
	}
}

int main(){
	scanf("%d",&n);
	for(int p,i=1;i<=n;i++){scanf("%d",&p);add(p,i);fa[i]=p;}
	int now=1;
	for(;!used[now];now=fa[now]) used[now]=1;
	for(;!cir[now];now=fa[now]) cir[now]=1;
	memset(used,0,sizeof(used));
	int num=0,maxi=-inf,mini=inf;
	for(int i=1;i<=n;i++)
		if(cir[i]){
			dfs(i);
			num++;
			maxi=max(maxi,seg[i]);
			mini=min(mini,seg[i]);
		}
	if(maxi==mini&&num%2) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}
```