#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 200005;
int n, fa[maxn], sg[maxn], pos = 1, tot, maxi = -1, mini = 0x3f3f3f3f;
bool vis[maxn], bucket[maxn];
vector<int> e[maxn];
void dfs(int u) {
	for(auto v : e[u]) 
		if(!vis[v]) dfs(v);
	for(auto v : e[u])
		if(!vis[v]) bucket[sg[v]] = true;
	while(bucket[sg[u]]) ++sg[u];
	for(auto v : e[u])
		if(!vis[v]) bucket[sg[v]] = false;
}
int main() {
	ios :: sync_with_stdio(false);
	cin >> n;
	for(int i = 1; i <= n; ++i)
		cin >> fa[i], e[fa[i]].push_back(i);
	while(!vis[pos]) vis[pos] = true, pos = fa[pos];
	memset(vis, 0, sizeof vis);
	while(!vis[pos]) vis[pos] = true, pos = fa[pos], ++tot;
	if(tot & 1) for(int i = 1; i <= n; ++i)
		if(vis[i]) dfs(i), maxi = max(maxi, sg[i]), mini = min(mini, sg[i]);
	puts(tot & 1 and maxi == mini ? "IMPOSSIBLE" : "POSSIBLE");
}
### Prompt

Your task is to create 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 <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 200005;
int n, fa[maxn], sg[maxn], pos = 1, tot, maxi = -1, mini = 0x3f3f3f3f;
bool vis[maxn], bucket[maxn];
vector<int> e[maxn];
void dfs(int u) {
	for(auto v : e[u]) 
		if(!vis[v]) dfs(v);
	for(auto v : e[u])
		if(!vis[v]) bucket[sg[v]] = true;
	while(bucket[sg[u]]) ++sg[u];
	for(auto v : e[u])
		if(!vis[v]) bucket[sg[v]] = false;
}
int main() {
	ios :: sync_with_stdio(false);
	cin >> n;
	for(int i = 1; i <= n; ++i)
		cin >> fa[i], e[fa[i]].push_back(i);
	while(!vis[pos]) vis[pos] = true, pos = fa[pos];
	memset(vis, 0, sizeof vis);
	while(!vis[pos]) vis[pos] = true, pos = fa[pos], ++tot;
	if(tot & 1) for(int i = 1; i <= n; ++i)
		if(vis[i]) dfs(i), maxi = max(maxi, sg[i]), mini = min(mini, sg[i]);
	puts(tot & 1 and maxi == mini ? "IMPOSSIBLE" : "POSSIBLE");
}
```