#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5+100;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

int head[N],nxt[N],p[N],st[N];
int vis[N],in[N],f[N],cnt=1;

inline int F(int k) {
	++cnt;
	for (int i=head[k];i;i=nxt[i])
		vis[f[i]]=cnt;
	for (int i=0;;i++)
		if (vis[i]!=cnt)
			return i;
}

inline void dfs(int k) {
	for (int i=head[k];i;i=nxt[i])
		dfs(i);
	f[k]=F(k);
}

int main()
{
	
	int n=gi(),i,k,top=0,a,b;
	
	for (i=1;i<=n;i++) p[i]=gi();
	for (k=1;!vis[k];k=p[k]) vis[k]=true;
	while (k!=st[1]) in[st[++top]=k]=true,k=p[k];
	for (k=1;k<=n;k++) if (!in[k]) nxt[k]=head[p[k]],head[p[k]]=k;
	
	for (i=1;i<=top;i++) dfs(st[i]);
	for (k=1;k<=n;k++) if (in[k]) nxt[k]=head[p[k]],head[p[k]]=k;
	for (i=0;vis[i]==cnt;i++);a=i;
	for (++i;vis[i]==cnt;i++);b=i;

	f[st[top]]=a;
	for (i=1;i<top;i++) f[st[i]]=F(st[i]);
	if (f[st[top-1]]!=a) return puts("POSSIBLE"),0;
	f[st[top]]=b;
	for (i=1;i<top;i++) f[st[i]]=F(st[i]);
	puts(f[st[top-1]]==a?"POSSIBLE":"IMPOSSIBLE");
	return 0;
}

### 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<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5+100;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

int head[N],nxt[N],p[N],st[N];
int vis[N],in[N],f[N],cnt=1;

inline int F(int k) {
	++cnt;
	for (int i=head[k];i;i=nxt[i])
		vis[f[i]]=cnt;
	for (int i=0;;i++)
		if (vis[i]!=cnt)
			return i;
}

inline void dfs(int k) {
	for (int i=head[k];i;i=nxt[i])
		dfs(i);
	f[k]=F(k);
}

int main()
{
	
	int n=gi(),i,k,top=0,a,b;
	
	for (i=1;i<=n;i++) p[i]=gi();
	for (k=1;!vis[k];k=p[k]) vis[k]=true;
	while (k!=st[1]) in[st[++top]=k]=true,k=p[k];
	for (k=1;k<=n;k++) if (!in[k]) nxt[k]=head[p[k]],head[p[k]]=k;
	
	for (i=1;i<=top;i++) dfs(st[i]);
	for (k=1;k<=n;k++) if (in[k]) nxt[k]=head[p[k]],head[p[k]]=k;
	for (i=0;vis[i]==cnt;i++);a=i;
	for (++i;vis[i]==cnt;i++);b=i;

	f[st[top]]=a;
	for (i=1;i<top;i++) f[st[i]]=F(st[i]);
	if (f[st[top-1]]!=a) return puts("POSSIBLE"),0;
	f[st[top]]=b;
	for (i=1;i<top;i++) f[st[i]]=F(st[i]);
	puts(f[st[top-1]]==a?"POSSIBLE":"IMPOSSIBLE");
	return 0;
}

```