#include<bits/stdc++.h>
using namespace std;
const int N=200005;
int ne[N],tot,fi[N],zz[N],Stack[N],cnt,vis[N],a[N],mex[N],x,n,flag[N];
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs1(int x){
	flag[x]=1;Stack[++cnt]=x;
	for (int i=fi[x];i;i=ne[i])
		if (!flag[zz[i]])dfs1(zz[i]);
		else if (flag[zz[i]]==1){
			tot=cnt;
			for (int i=1;i<=cnt;i++)a[i]=Stack[i],vis[a[i]]=1;
		}
	flag[x]=2;
	cnt--;
}
void dfs2(int x){
	for (int i=fi[x];i;i=ne[i])
		if (!vis[zz[i]])dfs2(zz[i]);
	for (int i=fi[x];i;i=ne[i])
		if (!vis[zz[i]])flag[mex[zz[i]]]=x;
	while (flag[mex[x]]==x)mex[x]++;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&x);
		jb(x,i);
	}
	tot=0;
	for (int i=1;i<=n;i++)
		if (!flag[i])dfs1(i);
	if (tot%2==0){
		puts("POSSIBLE");
		return 0;
	}
	memset(flag,0,sizeof flag);
	for (int i=1;i<=tot;i++)dfs2(a[i]);
	for (int i=2;i<=tot;i++)
		if (mex[a[i]]!=mex[a[1]]){
			puts("POSSIBLE");
			return 0;
		}
	puts("IMPOSSIBLE");
}

### Prompt

Your challenge is to write 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;
const int N=200005;
int ne[N],tot,fi[N],zz[N],Stack[N],cnt,vis[N],a[N],mex[N],x,n,flag[N];
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs1(int x){
	flag[x]=1;Stack[++cnt]=x;
	for (int i=fi[x];i;i=ne[i])
		if (!flag[zz[i]])dfs1(zz[i]);
		else if (flag[zz[i]]==1){
			tot=cnt;
			for (int i=1;i<=cnt;i++)a[i]=Stack[i],vis[a[i]]=1;
		}
	flag[x]=2;
	cnt--;
}
void dfs2(int x){
	for (int i=fi[x];i;i=ne[i])
		if (!vis[zz[i]])dfs2(zz[i]);
	for (int i=fi[x];i;i=ne[i])
		if (!vis[zz[i]])flag[mex[zz[i]]]=x;
	while (flag[mex[x]]==x)mex[x]++;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&x);
		jb(x,i);
	}
	tot=0;
	for (int i=1;i<=n;i++)
		if (!flag[i])dfs1(i);
	if (tot%2==0){
		puts("POSSIBLE");
		return 0;
	}
	memset(flag,0,sizeof flag);
	for (int i=1;i<=tot;i++)dfs2(a[i]);
	for (int i=2;i<=tot;i++)
		if (mex[a[i]]!=mex[a[1]]){
			puts("POSSIBLE");
			return 0;
		}
	puts("IMPOSSIBLE");
}

```