#include <bits/stdc++.h>
using namespace std;

#define fo(i,s,t) for(int i = s; i <= t; ++ i)
#define fd(i,s,t) for(int i = s; i >= t; -- i)
typedef long long ll;
#define pb push_back

const int maxn = 200050;
const int inf = 1<<30;

int n, in[maxn], to[maxn];
bool v[maxn];
set<int> s[maxn];
vector<int> loop, candidate;

void go(int x)
{
	v[x] = true;
	fo(i,0,inf) if(s[x].find(i)==s[x].end()){s[to[x]].insert(i); break;}
	if(!(-- in[to[x]])) go(to[x]);
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n) {int x; scanf("%d",&to[i]); in[to[i]] ++;}
	fo(i,1,n) if(!in[i] && !v[i]) go(i);
	fo(i,1,n) if(!v[i]) 
	{
		int now = i; 
		while(1)
		{
			loop.pb(now);
			now = to[now];
			if(now == i) break;
		}
		break;
	}
	for(int i = 0; candidate.size()<2; ++ i)
		if(s[loop[0]].find(i)==s[loop[0]].end())
			candidate.pb(i);
	fo(i,0,1)
	{
		in[0] = candidate[i];
		fo(j,1,loop.size()-1)
		{
			fo(k,0,inf)
				if(in[j-1]!=k&&s[loop[j]].find(k)==s[loop[j]].end())
				{
					in[j] = k;
					break;
				}
		}
		if(i==0&&in[loop.size()-1]!=in[0]) return 0*printf("POSSIBLE\n");
		if(i&&in[loop.size()-1]==candidate[0]) return 0*printf("POSSIBLE\n");
	}
	printf("IMPOSSIBLE\n");
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 fo(i,s,t) for(int i = s; i <= t; ++ i)
#define fd(i,s,t) for(int i = s; i >= t; -- i)
typedef long long ll;
#define pb push_back

const int maxn = 200050;
const int inf = 1<<30;

int n, in[maxn], to[maxn];
bool v[maxn];
set<int> s[maxn];
vector<int> loop, candidate;

void go(int x)
{
	v[x] = true;
	fo(i,0,inf) if(s[x].find(i)==s[x].end()){s[to[x]].insert(i); break;}
	if(!(-- in[to[x]])) go(to[x]);
}
int main()
{
	scanf("%d",&n);
	fo(i,1,n) {int x; scanf("%d",&to[i]); in[to[i]] ++;}
	fo(i,1,n) if(!in[i] && !v[i]) go(i);
	fo(i,1,n) if(!v[i]) 
	{
		int now = i; 
		while(1)
		{
			loop.pb(now);
			now = to[now];
			if(now == i) break;
		}
		break;
	}
	for(int i = 0; candidate.size()<2; ++ i)
		if(s[loop[0]].find(i)==s[loop[0]].end())
			candidate.pb(i);
	fo(i,0,1)
	{
		in[0] = candidate[i];
		fo(j,1,loop.size()-1)
		{
			fo(k,0,inf)
				if(in[j-1]!=k&&s[loop[j]].find(k)==s[loop[j]].end())
				{
					in[j] = k;
					break;
				}
		}
		if(i==0&&in[loop.size()-1]!=in[0]) return 0*printf("POSSIBLE\n");
		if(i&&in[loop.size()-1]==candidate[0]) return 0*printf("POSSIBLE\n");
	}
	printf("IMPOSSIBLE\n");
	return 0;
}

```