#include <cstdio>
#include <cctype>
#include <cstring>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repe(i,x) for (edge *i=fst[x]; i; i=i->nxt)
using namespace std;

int n,p[200100],root,f[200100];
bool cyc[200100];
struct edge
{
    int v;
    edge *nxt;
} pool[200100],*tp=pool,*fst[200100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

int mex(int x,bool type)
{
    static bool flag[200100];
    repe(i,x)
        if (type || !cyc[i->v])
            flag[f[i->v]]=1;
    int ret=0;
    for (; flag[ret]; ++ret);
    repe(i,x)
        if (type || !cyc[i->v])
            flag[f[i->v]]=0;
    return ret;
}

void dfs(int x)
{
    repe(i,x)
        if (!cyc[i->v])
            dfs(i->v);
    f[x]=mex(x,0);
}

int main()
{
    n=getint();
    repu(i,1,n)
    {
        p[i]=getint();
        *tp=(edge){i,fst[p[i]]},fst[p[i]]=tp++;
    }
    root=1;
    for (; !cyc[root]; cyc[root]=1,root=p[root]);
    memset(cyc,0,sizeof(cyc));
    for (; !cyc[root]; cyc[root]=1,root=p[root]);
    repu(i,1,n)
        if (cyc[i])
            dfs(i);
    for (int i=p[p[root]]; i!=p[root]; f[i]=mex(i,1),i=p[i]);
    if (f[root]!=f[p[root]])
    {
        puts("POSSIBLE");
        return 0;
    }
    ++f[p[root]];
    for (int i=p[p[root]]; i!=p[root]; f[i]=mex(i,1),i=p[i]);
    puts(f[root]+1==f[p[root]]?"POSSIBLE":"IMPOSSIBLE");
    return 0;
}
### 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 <cstdio>
#include <cctype>
#include <cstring>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repe(i,x) for (edge *i=fst[x]; i; i=i->nxt)
using namespace std;

int n,p[200100],root,f[200100];
bool cyc[200100];
struct edge
{
    int v;
    edge *nxt;
} pool[200100],*tp=pool,*fst[200100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

int mex(int x,bool type)
{
    static bool flag[200100];
    repe(i,x)
        if (type || !cyc[i->v])
            flag[f[i->v]]=1;
    int ret=0;
    for (; flag[ret]; ++ret);
    repe(i,x)
        if (type || !cyc[i->v])
            flag[f[i->v]]=0;
    return ret;
}

void dfs(int x)
{
    repe(i,x)
        if (!cyc[i->v])
            dfs(i->v);
    f[x]=mex(x,0);
}

int main()
{
    n=getint();
    repu(i,1,n)
    {
        p[i]=getint();
        *tp=(edge){i,fst[p[i]]},fst[p[i]]=tp++;
    }
    root=1;
    for (; !cyc[root]; cyc[root]=1,root=p[root]);
    memset(cyc,0,sizeof(cyc));
    for (; !cyc[root]; cyc[root]=1,root=p[root]);
    repu(i,1,n)
        if (cyc[i])
            dfs(i);
    for (int i=p[p[root]]; i!=p[root]; f[i]=mex(i,1),i=p[i]);
    if (f[root]!=f[p[root]])
    {
        puts("POSSIBLE");
        return 0;
    }
    ++f[p[root]];
    for (int i=p[p[root]]; i!=p[root]; f[i]=mex(i,1),i=p[i]);
    puts(f[root]+1==f[p[root]]?"POSSIBLE":"IMPOSSIBLE");
    return 0;
}
```