#include<cstdio>
#include<vector>
#include<cstring>
#define rep(i, a, b) for (register int i=(a); i<=(b); ++i)
#define per(i, a, b) for (register int i=(a); i>=(b); --i)
using namespace std;
const int N=200005;
int cir[N], vis[N], mex[N], fa[N];
vector<int> G[N];

inline int read()
{
 	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}

void dfs(int u)
{
	for (int v: G[u]) if (!cir[v]) dfs(v);
	for (int v: G[u]) if (!cir[v]) vis[mex[v]]=1;
	while (vis[mex[u]]) mex[u]++;
	for (int v: G[u]) if (!cir[v]) vis[mex[v]]=0;
}

int main()
{
	int n=read(), now=1;
	rep(i, 1, n) G[fa[i]=read()].push_back(i);
	while (!cir[now]) cir[now]=1, now=fa[now];
	memset(cir, 0, sizeof(cir));
	while (!cir[now]) cir[now]=1, now=fa[now];
	int Max=-1, Min=n+1, cnt=0;
	rep(i, 1, n) if (cir[i])
	{
		dfs(i); cnt++;
		Max=max(Max, mex[i]);
		Min=min(Min, mex[i]);
	}
	puts(Max==Min&&(cnt&1)?"IMPOSSIBLE":"POSSIBLE");
	return 0;
}

### Prompt

Develop a solution in cpp 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<vector>
#include<cstring>
#define rep(i, a, b) for (register int i=(a); i<=(b); ++i)
#define per(i, a, b) for (register int i=(a); i>=(b); --i)
using namespace std;
const int N=200005;
int cir[N], vis[N], mex[N], fa[N];
vector<int> G[N];

inline int read()
{
 	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}

void dfs(int u)
{
	for (int v: G[u]) if (!cir[v]) dfs(v);
	for (int v: G[u]) if (!cir[v]) vis[mex[v]]=1;
	while (vis[mex[u]]) mex[u]++;
	for (int v: G[u]) if (!cir[v]) vis[mex[v]]=0;
}

int main()
{
	int n=read(), now=1;
	rep(i, 1, n) G[fa[i]=read()].push_back(i);
	while (!cir[now]) cir[now]=1, now=fa[now];
	memset(cir, 0, sizeof(cir));
	while (!cir[now]) cir[now]=1, now=fa[now];
	int Max=-1, Min=n+1, cnt=0;
	rep(i, 1, n) if (cir[i])
	{
		dfs(i); cnt++;
		Max=max(Max, mex[i]);
		Min=min(Min, mex[i]);
	}
	puts(Max==Min&&(cnt&1)?"IMPOSSIBLE":"POSSIBLE");
	return 0;
}

```