// Code by ajcxsu
// Problem: namori grundy

#include<bits/stdc++.h>
using namespace std;

const int N=2e5+10;
int h[N], to[N], nexp[N], p=1;
inline void ins(int a, int b) { nexp[p]=h[a], h[a]=p, to[p]=b, p++; }

char tak[N], vis[N], viis[N];
int mex[N], smex[N], cir[N], nd[N], fa[N], idx, flag;
void dfs2(int x) {
    int siz=0;
    for(int u=h[x];u;u=nexp[u])
        if(!cir[to[u]]) dfs2(to[u]), ++siz;
    fill(tak, tak+siz+5, 0);
    for(int u=h[x];u;u=nexp[u])
        if(!cir[to[u]]) tak[mex[to[u]]]=1;
    for(mex[x]=0;tak[mex[x]];mex[x]++);
    for(smex[x]=mex[x]+1;tak[smex[x]];smex[x]++);
}

int rm[N];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n;
    cin>>n;
    int u;
    for(int i=1;i<=n;i++) cin>>fa[i], ins(fa[i], i);
    int na=1, rt;
    while(!vis[na]) vis[na]=1, na=fa[na];
    rt=na;
    do { cir[na]=++idx, nd[idx]=na, na=fa[na]; } while(rt!=na);
    for(int i=1;i<=idx;i++) dfs2(nd[i]);
    rm[1]=mex[nd[1]];
    for(int i=2;i<=idx;i++)
        rm[i]=(rm[i-1]==mex[nd[i]]?smex[nd[i]]:mex[nd[i]]);
    if(rm[idx]!=rm[1]) cout<<"POSSIBLE"<<endl, exit(0);
    rm[1]=smex[nd[1]];
    for(int i=2;i<=idx;i++)
        rm[i]=(rm[i-1]==mex[nd[i]]?smex[nd[i]]:mex[nd[i]]);
    // for(int i=1;i<=idx;i++) cout<<mex[i]<<" \n"[i==idx];
    if(rm[idx]==mex[nd[1]]) cout<<"POSSIBLE"<<endl;
    else cout<<"IMPOSSIBLE"<<endl;
    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
// Code by ajcxsu
// Problem: namori grundy

#include<bits/stdc++.h>
using namespace std;

const int N=2e5+10;
int h[N], to[N], nexp[N], p=1;
inline void ins(int a, int b) { nexp[p]=h[a], h[a]=p, to[p]=b, p++; }

char tak[N], vis[N], viis[N];
int mex[N], smex[N], cir[N], nd[N], fa[N], idx, flag;
void dfs2(int x) {
    int siz=0;
    for(int u=h[x];u;u=nexp[u])
        if(!cir[to[u]]) dfs2(to[u]), ++siz;
    fill(tak, tak+siz+5, 0);
    for(int u=h[x];u;u=nexp[u])
        if(!cir[to[u]]) tak[mex[to[u]]]=1;
    for(mex[x]=0;tak[mex[x]];mex[x]++);
    for(smex[x]=mex[x]+1;tak[smex[x]];smex[x]++);
}

int rm[N];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n;
    cin>>n;
    int u;
    for(int i=1;i<=n;i++) cin>>fa[i], ins(fa[i], i);
    int na=1, rt;
    while(!vis[na]) vis[na]=1, na=fa[na];
    rt=na;
    do { cir[na]=++idx, nd[idx]=na, na=fa[na]; } while(rt!=na);
    for(int i=1;i<=idx;i++) dfs2(nd[i]);
    rm[1]=mex[nd[1]];
    for(int i=2;i<=idx;i++)
        rm[i]=(rm[i-1]==mex[nd[i]]?smex[nd[i]]:mex[nd[i]]);
    if(rm[idx]!=rm[1]) cout<<"POSSIBLE"<<endl, exit(0);
    rm[1]=smex[nd[1]];
    for(int i=2;i<=idx;i++)
        rm[i]=(rm[i-1]==mex[nd[i]]?smex[nd[i]]:mex[nd[i]]);
    // for(int i=1;i<=idx;i++) cout<<mex[i]<<" \n"[i==idx];
    if(rm[idx]==mex[nd[1]]) cout<<"POSSIBLE"<<endl;
    else cout<<"IMPOSSIBLE"<<endl;
    return 0;
}

```