#include<bits/stdc++.h>
using namespace std;
int fa[1000100],st[1000100],a[1000100],js[1000100],tim;
bool vis[1000100];
vector <int> des[1000100];
void dfs(int s)
{
	for (int k=0;k<des[s].size();k++) if (!vis[des[s][k]]) dfs(des[s][k]);
	tim++;
	for (int k=0;k<des[s].size();k++) if (!vis[des[s][k]]) js[a[des[s][k]]]=tim;
	for (a[s]=0;;a[s]++) if (js[a[s]]!=tim) break;
}
int main()
{
	int n,top=0;scanf("%d",&n);
	for (int i=1;i<=n;i++) {scanf("%d",&fa[i]);des[fa[i]].push_back(i);}
	int now=1;while (!vis[now]) {st[++top]=now;vis[now]=true;now=fa[now];}
	int p=1;while (st[p]!=now) p++;st[top+1]=st[p];
	if ((top-p)&1) {printf("POSSIBLE\n");return 0;}
	memset(vis,0,sizeof(vis));
	for (int i=p;i<=top;i++) vis[st[i]]=true;
	for (int i=p;i<=top;i++) dfs(st[i]);
	for (int i=p;i<=top;i++) if (a[st[i]]!=a[st[i+1]]) {printf("POSSIBLE\n");return 0;}
	printf("IMPOSSIBLE\n");
}
### Prompt

In CPP, your task is to solve 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>
using namespace std;
int fa[1000100],st[1000100],a[1000100],js[1000100],tim;
bool vis[1000100];
vector <int> des[1000100];
void dfs(int s)
{
	for (int k=0;k<des[s].size();k++) if (!vis[des[s][k]]) dfs(des[s][k]);
	tim++;
	for (int k=0;k<des[s].size();k++) if (!vis[des[s][k]]) js[a[des[s][k]]]=tim;
	for (a[s]=0;;a[s]++) if (js[a[s]]!=tim) break;
}
int main()
{
	int n,top=0;scanf("%d",&n);
	for (int i=1;i<=n;i++) {scanf("%d",&fa[i]);des[fa[i]].push_back(i);}
	int now=1;while (!vis[now]) {st[++top]=now;vis[now]=true;now=fa[now];}
	int p=1;while (st[p]!=now) p++;st[top+1]=st[p];
	if ((top-p)&1) {printf("POSSIBLE\n");return 0;}
	memset(vis,0,sizeof(vis));
	for (int i=p;i<=top;i++) vis[st[i]]=true;
	for (int i=p;i<=top;i++) dfs(st[i]);
	for (int i=p;i<=top;i++) if (a[st[i]]!=a[st[i+1]]) {printf("POSSIBLE\n");return 0;}
	printf("IMPOSSIBLE\n");
}
```