#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n,mx,mn;
int p[200010];
bool vis[200010],cir[200010],_is[200010];
int mex[200010],cnt_tree=0;
struct node{int y,nxt;
}g[200010];int h[200010],tot=0;
void dfs(int x)
{
    for (int i=h[x];i;i=g[i].nxt) if (!cir[g[i].y]) dfs(g[i].y);
    for (int i=h[x];i;i=g[i].nxt) if (!cir[g[i].y]) _is[mex[g[i].y]]=1;
    for (mex[x]=0;_is[mex[x]];mex[x]++);
    for (int i=h[x];i;i=g[i].nxt) _is[mex[g[i].y]]=0;
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) 
    {
        scanf("%d",&p[i]);
        g[++tot]=node{i,h[p[i]]};h[p[i]]=i;
    }
    int x=1;
    while (!vis[x]) vis[x]=1,x=p[x];
    while (!cir[x]) cir[x]=1,x=p[x];
    mx=0;mn=2147483647;
    for (int i=1;i<=n;i++)
    if (cir[i])
    {
        dfs(i);cnt_tree++;
        mx=max(mx,mex[i]);mn=min(mn,mex[i]);
    }
    if (mx==mn&&(cnt_tree&1)) printf("IMPOSSIBLE");else printf("POSSIBLE");
}
### 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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n,mx,mn;
int p[200010];
bool vis[200010],cir[200010],_is[200010];
int mex[200010],cnt_tree=0;
struct node{int y,nxt;
}g[200010];int h[200010],tot=0;
void dfs(int x)
{
    for (int i=h[x];i;i=g[i].nxt) if (!cir[g[i].y]) dfs(g[i].y);
    for (int i=h[x];i;i=g[i].nxt) if (!cir[g[i].y]) _is[mex[g[i].y]]=1;
    for (mex[x]=0;_is[mex[x]];mex[x]++);
    for (int i=h[x];i;i=g[i].nxt) _is[mex[g[i].y]]=0;
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) 
    {
        scanf("%d",&p[i]);
        g[++tot]=node{i,h[p[i]]};h[p[i]]=i;
    }
    int x=1;
    while (!vis[x]) vis[x]=1,x=p[x];
    while (!cir[x]) cir[x]=1,x=p[x];
    mx=0;mn=2147483647;
    for (int i=1;i<=n;i++)
    if (cir[i])
    {
        dfs(i);cnt_tree++;
        mx=max(mx,mex[i]);mn=min(mn,mex[i]);
    }
    if (mx==mn&&(cnt_tree&1)) printf("IMPOSSIBLE");else printf("POSSIBLE");
}
```