#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int N = 2e5 + 5; 
using namespace std; 

int n, fa[N], head[N], cnte, vis[N], cir[N], mex[N];
struct edge { int to, nxt; } e[N << 1]; 

template < typename T >
inline T read()
{
	T 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; 
}

inline void adde(int u, int v) { e[++cnte] = (edge) { v, head[u] }, head[u] = cnte; }

void dfs(int u)
{
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) dfs(e[i].to);
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) vis[mex[e[i].to]] = 1;
	int x = 0;
	while(vis[x]) x++;
	mex[u] = x;
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) vis[mex[e[i].to]] = 0; 
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("cpp.in", "r", stdin); 
#endif
	n = read <int> ();
	for(int i = 1; i <= n; i++) fa[i] = read <int> (), adde(fa[i], i);
	int u = 1;
	while(!vis[u]) vis[u] = 1, u = fa[u];
	for(int i = 0; i <= n; i++) vis[i] = 0; 
	while(!cir[u]) cir[u] = 1, u = fa[u];
	int mn = n + 1, mx = 0, cnt = 0;
	for(int i = 1; i <= n; i++)
	{
		if(!cir[i]) continue;
		dfs(i), cnt++;
		mx = max(mx, mex[i]), mn = min(mn, mex[i]); 
	}
	puts((mx == mn) && (cnt & 1) ? "IMPOSSIBLE" : "POSSIBLE"); 
	return 0; 
}

### Prompt

Please create a solution in Cpp 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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int N = 2e5 + 5; 
using namespace std; 

int n, fa[N], head[N], cnte, vis[N], cir[N], mex[N];
struct edge { int to, nxt; } e[N << 1]; 

template < typename T >
inline T read()
{
	T 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; 
}

inline void adde(int u, int v) { e[++cnte] = (edge) { v, head[u] }, head[u] = cnte; }

void dfs(int u)
{
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) dfs(e[i].to);
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) vis[mex[e[i].to]] = 1;
	int x = 0;
	while(vis[x]) x++;
	mex[u] = x;
	for(int i = head[u]; i; i = e[i].nxt)
		if(!cir[e[i].to]) vis[mex[e[i].to]] = 0; 
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("cpp.in", "r", stdin); 
#endif
	n = read <int> ();
	for(int i = 1; i <= n; i++) fa[i] = read <int> (), adde(fa[i], i);
	int u = 1;
	while(!vis[u]) vis[u] = 1, u = fa[u];
	for(int i = 0; i <= n; i++) vis[i] = 0; 
	while(!cir[u]) cir[u] = 1, u = fa[u];
	int mn = n + 1, mx = 0, cnt = 0;
	for(int i = 1; i <= n; i++)
	{
		if(!cir[i]) continue;
		dfs(i), cnt++;
		mx = max(mx, mex[i]), mn = min(mn, mex[i]); 
	}
	puts((mx == mn) && (cnt & 1) ? "IMPOSSIBLE" : "POSSIBLE"); 
	return 0; 
}

```