#include<bits/stdc++.h>
#define maxn 400005
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);i>=LIM;i--)
using namespace std;

int n,fa[maxn];
int info[maxn],Prev[maxn<<1],to[maxn<<1],cnt_e=1;
void Node(int u,int v){ Prev[++cnt_e]=info[u],info[u]=cnt_e,to[cnt_e]=v; }

int ar[maxn],onc[maxn],a[maxn];
int vis[maxn],sta[maxn],tp;
void dfs0(int u,int ff){
	vis[u] = 1;
	sta[++tp] = u;
	for(int i=info[u],v;i;i=Prev[i]) if(i^ff){
		v=to[i];
		if(!vis[v]) dfs0(v,(i^1));
		else if(!onc[v]){
			for(int t=-1;t!=v;){
				t = sta[tp--];
				ar[++ar[0]] = t;
				onc[t] = 1;
			}
			per(i,ar[0],1) sta[++tp] = ar[i];
		}
	}
	if(sta[tp] == u) tp--;
}

void dfs1(int u,int ff){
	static int vis[maxn],tim;
	for(int i=info[u],v;i;i=Prev[i]) if((v=to[i]) ^ ff && !onc[v])
		dfs1(v,u);
	++tim;
	for(int i=info[u],v;i;i=Prev[i]) if((v=to[i]) ^ ff && !onc[v])
		vis[a[v]] = tim;
	a[u] = 0;
	for(;vis[a[u]] == tim; a[u]++);
}

int main(){
	scanf("%d",&n);
	rep(i,1,n){
		scanf("%d",&fa[i]);
		Node(fa[i],i),Node(i,fa[i]);
	}
	dfs0(1,0);
	rep(i,1,ar[0]) dfs1(ar[i],0);
	set<int>st;
	rep(i,1,ar[0]) st.insert(a[ar[i]]);
	if((ar[0] & 1) && st.size() == 1)
		puts("IMPOSSIBLE");
	else 
		puts("POSSIBLE");
}
### Prompt

Please formulate 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>
#define maxn 400005
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);i>=LIM;i--)
using namespace std;

int n,fa[maxn];
int info[maxn],Prev[maxn<<1],to[maxn<<1],cnt_e=1;
void Node(int u,int v){ Prev[++cnt_e]=info[u],info[u]=cnt_e,to[cnt_e]=v; }

int ar[maxn],onc[maxn],a[maxn];
int vis[maxn],sta[maxn],tp;
void dfs0(int u,int ff){
	vis[u] = 1;
	sta[++tp] = u;
	for(int i=info[u],v;i;i=Prev[i]) if(i^ff){
		v=to[i];
		if(!vis[v]) dfs0(v,(i^1));
		else if(!onc[v]){
			for(int t=-1;t!=v;){
				t = sta[tp--];
				ar[++ar[0]] = t;
				onc[t] = 1;
			}
			per(i,ar[0],1) sta[++tp] = ar[i];
		}
	}
	if(sta[tp] == u) tp--;
}

void dfs1(int u,int ff){
	static int vis[maxn],tim;
	for(int i=info[u],v;i;i=Prev[i]) if((v=to[i]) ^ ff && !onc[v])
		dfs1(v,u);
	++tim;
	for(int i=info[u],v;i;i=Prev[i]) if((v=to[i]) ^ ff && !onc[v])
		vis[a[v]] = tim;
	a[u] = 0;
	for(;vis[a[u]] == tim; a[u]++);
}

int main(){
	scanf("%d",&n);
	rep(i,1,n){
		scanf("%d",&fa[i]);
		Node(fa[i],i),Node(i,fa[i]);
	}
	dfs0(1,0);
	rep(i,1,ar[0]) dfs1(ar[i],0);
	set<int>st;
	rep(i,1,ar[0]) st.insert(a[ar[i]]);
	if((ar[0] & 1) && st.size() == 1)
		puts("IMPOSSIBLE");
	else 
		puts("POSSIBLE");
}
```