#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define pb push_back

using namespace std;

const int N = 2e5 + 10;

int n;
int p[N], fa[N];

int find(int x){
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int rt, ban;
int val[N];
vector<int> G[N];

bool vis[N];

void DFS(int o){
	if(o == ban) return;
	if(!G[o].size()) return;
	for(int v : G[o]) DFS(v);
	for(int v : G[o]) vis[val[v]] = true;
	For(i, 0, n) if(!vis[i]){
		val[o] = i; 
		break;
	} 
	for(int v : G[o]) vis[val[v]] = false;
}

int main(){
	scanf("%d", &n);
	For(i, 1, n) fa[i] = i;
	For(i, 1, n){
		scanf("%d", &p[i]);
		if(find(i) == find(p[i])) rt = i;
		else fa[find(i)] = find(p[i]), G[p[i]].pb(i);
	}
	DFS(rt);
	if(val[rt] == val[p[rt]]){
		ban = p[rt];
		val[p[rt]]++;
		DFS(rt);
		if(val[rt] != val[p[rt]] - 1){
			puts("IMPOSSIBLE");
			return 0;
		}
	}
	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
#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define pb push_back

using namespace std;

const int N = 2e5 + 10;

int n;
int p[N], fa[N];

int find(int x){
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int rt, ban;
int val[N];
vector<int> G[N];

bool vis[N];

void DFS(int o){
	if(o == ban) return;
	if(!G[o].size()) return;
	for(int v : G[o]) DFS(v);
	for(int v : G[o]) vis[val[v]] = true;
	For(i, 0, n) if(!vis[i]){
		val[o] = i; 
		break;
	} 
	for(int v : G[o]) vis[val[v]] = false;
}

int main(){
	scanf("%d", &n);
	For(i, 1, n) fa[i] = i;
	For(i, 1, n){
		scanf("%d", &p[i]);
		if(find(i) == find(p[i])) rt = i;
		else fa[find(i)] = find(p[i]), G[p[i]].pb(i);
	}
	DFS(rt);
	if(val[rt] == val[p[rt]]){
		ban = p[rt];
		val[p[rt]]++;
		DFS(rt);
		if(val[rt] != val[p[rt]] - 1){
			puts("IMPOSSIBLE");
			return 0;
		}
	}
	puts("POSSIBLE");
	return 0;
}
```