// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=200000+10;

int n,fa[N],vis[N],inc[N],f[N],o[N];
vector<int> E[N];

void dfs(int u) {
	for (int v:E[u]) if (!inc[v]) dfs(v);
	for (int v:E[u]) if (!inc[v]) o[f[v]]=1;
	while (o[f[u]]) ++f[u];
	for (int v:E[u]) if (!inc[v]) o[f[v]]=0;
}

int main() {
	n=read();
	for (int i=1;i<=n;++i)
		E[fa[i]=read()].emplace_back(i);
	int x=1;
	for (;!vis[x];x=fa[x]) vis[x]=1;
	for (;!inc[x];x=fa[x]) inc[x]=1;
	int cnt=0,mn=1e9,mx=-1e9;
	for (int i=1;i<=n;++i)
		if (inc[i]) dfs(i),mn=min(mn,f[i]),mx=max(mx,f[i]),++cnt;
	if (mn==mx&&(cnt&1)) puts("IMPOSSIBLE");
	else puts("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
// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=200000+10;

int n,fa[N],vis[N],inc[N],f[N],o[N];
vector<int> E[N];

void dfs(int u) {
	for (int v:E[u]) if (!inc[v]) dfs(v);
	for (int v:E[u]) if (!inc[v]) o[f[v]]=1;
	while (o[f[u]]) ++f[u];
	for (int v:E[u]) if (!inc[v]) o[f[v]]=0;
}

int main() {
	n=read();
	for (int i=1;i<=n;++i)
		E[fa[i]=read()].emplace_back(i);
	int x=1;
	for (;!vis[x];x=fa[x]) vis[x]=1;
	for (;!inc[x];x=fa[x]) inc[x]=1;
	int cnt=0,mn=1e9,mx=-1e9;
	for (int i=1;i<=n;++i)
		if (inc[i]) dfs(i),mn=min(mn,f[i]),mx=max(mx,f[i]),++cnt;
	if (mn==mx&&(cnt&1)) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}
```