#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const double pi=acos(-1.0);
const int mod=1e9+7;
//LL powmod(LL a,LL b) {LL res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int p[200010],v[200010],c[200010],a[200010];
VI g[200010],e[200010];
int dfs(int x){
    for(auto i:g[x]){
        if(c[i]) continue;
        e[x].pb(dfs(i));
    }
    sort(all(e[x]));
    a[x]=-1;
    for(auto i:e[x]){
        if(i>a[x]+1) break;
        a[x]=i;
    }
    return ++a[x];
}
int main(){
//    freopen("out.txt","r",stdin);
//    freopen("outt.txt","w",stdout);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>p[i];
        g[p[i]].pb(i);
    }
    int r=1;
    while(!v[r]){
        v[r]=1;
        r=p[r];
    }
    while(!c[r]){
        c[r]=1;
        r=p[r];
    }
    for(int i=1;i<=n;i++){
        if(c[i]) dfs(i);
    }
    int f=0,cnt=0;
    while(c[r]){
        c[r]=0;
        if(a[r]!=a[p[r]]) f=1;
        r=p[r];
        cnt++;
    }
    puts((f||cnt%2==0)?"POSSIBLE" : "IMPOSSIBLE");
    return 0;
}

### Prompt

Create a solution in cpp for 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;
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const double pi=acos(-1.0);
const int mod=1e9+7;
//LL powmod(LL a,LL b) {LL res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int p[200010],v[200010],c[200010],a[200010];
VI g[200010],e[200010];
int dfs(int x){
    for(auto i:g[x]){
        if(c[i]) continue;
        e[x].pb(dfs(i));
    }
    sort(all(e[x]));
    a[x]=-1;
    for(auto i:e[x]){
        if(i>a[x]+1) break;
        a[x]=i;
    }
    return ++a[x];
}
int main(){
//    freopen("out.txt","r",stdin);
//    freopen("outt.txt","w",stdout);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>p[i];
        g[p[i]].pb(i);
    }
    int r=1;
    while(!v[r]){
        v[r]=1;
        r=p[r];
    }
    while(!c[r]){
        c[r]=1;
        r=p[r];
    }
    for(int i=1;i<=n;i++){
        if(c[i]) dfs(i);
    }
    int f=0,cnt=0;
    while(c[r]){
        c[r]=0;
        if(a[r]!=a[p[r]]) f=1;
        r=p[r];
        cnt++;
    }
    puts((f||cnt%2==0)?"POSSIBLE" : "IMPOSSIBLE");
    return 0;
}

```