#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, a[200005], t, rt, ln, rn;
bool used[200005], s;
vector<int> ch[200005];
int dfs(int t){
	int ret = 0;
	bool ap[20] = {};
	for(int u: ch[t]) if(u != rt) ap[dfs(u)] = true;
	if(t == a[rt] && s) ap[ln] = true;
	for(; ap[ret]; ret++);
	if(t == a[rt]) ln = ret;
	return ret;
}
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", a+i);
		ch[--a[i]].push_back(i);
	}
	for(; !used[t]; t = a[t]) used[t] = true;
	rt = t;
	rn = dfs(rt);
	if(rn == ln){
		s = true;
		if(rn != dfs(rt)) printf("IM");
	}
	puts("POSSIBLE");
}
### 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;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, a[200005], t, rt, ln, rn;
bool used[200005], s;
vector<int> ch[200005];
int dfs(int t){
	int ret = 0;
	bool ap[20] = {};
	for(int u: ch[t]) if(u != rt) ap[dfs(u)] = true;
	if(t == a[rt] && s) ap[ln] = true;
	for(; ap[ret]; ret++);
	if(t == a[rt]) ln = ret;
	return ret;
}
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", a+i);
		ch[--a[i]].push_back(i);
	}
	for(; !used[t]; t = a[t]) used[t] = true;
	rt = t;
	rn = dfs(rt);
	if(rn == ln){
		s = true;
		if(rn != dfs(rt)) printf("IM");
	}
	puts("POSSIBLE");
}
```