#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=200005;

int n,cnt,last[N],p[N],sg[N],p1,p2,tim,arr[N];
bool vis[N];
struct edge{int to,next;bool del;}e[N];

int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

void addedge(int u,int v)
{
    e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
}

int get_mex(int x)
{
    int ans;
    for (int i=last[x];i;i=e[i].next) if (!e[i].del) vis[sg[e[i].to]]=1;
    for (int i=0;;i++) if (!vis[i]) {ans=i;break;}
    for (int i=last[x];i;i=e[i].next) if (!e[i].del) vis[sg[e[i].to]]=0;
    return ans;
}

void dfs(int x)
{
    arr[x]=tim;
    for (int i=last[x];i;i=e[i].next)
        if (!arr[e[i].to]) dfs(e[i].to);
        else if (arr[e[i].to]==tim) p1=x,p2=e[i].to,e[i].del=1;
    sg[x]=get_mex(x);
}

int main()
{
    n=read();
    for (int i=1;i<=n;i++) p[i]=read(),addedge(p[i],i);
    for (int i=1;i<=n;i++) if (!arr[i]) tim++,dfs(i);
    int s1=get_mex(p1);
    vis[s1]=1;
    int s2=get_mex(p1);
    vis[s1]=0;
    if (sg[p2]!=s1) {puts("POSSIBLE");return 0;}
    sg[p1]=s2;
    int x=p1;
    while (x!=p2) x=p[x],sg[x]=get_mex(x);
    if (sg[p2]==s1) puts("POSSIBLE");
    else puts("IMPOSSIBLE");
    return 0;
}
### Prompt

Please create a solution in Cpp to 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<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=200005;

int n,cnt,last[N],p[N],sg[N],p1,p2,tim,arr[N];
bool vis[N];
struct edge{int to,next;bool del;}e[N];

int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

void addedge(int u,int v)
{
    e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
}

int get_mex(int x)
{
    int ans;
    for (int i=last[x];i;i=e[i].next) if (!e[i].del) vis[sg[e[i].to]]=1;
    for (int i=0;;i++) if (!vis[i]) {ans=i;break;}
    for (int i=last[x];i;i=e[i].next) if (!e[i].del) vis[sg[e[i].to]]=0;
    return ans;
}

void dfs(int x)
{
    arr[x]=tim;
    for (int i=last[x];i;i=e[i].next)
        if (!arr[e[i].to]) dfs(e[i].to);
        else if (arr[e[i].to]==tim) p1=x,p2=e[i].to,e[i].del=1;
    sg[x]=get_mex(x);
}

int main()
{
    n=read();
    for (int i=1;i<=n;i++) p[i]=read(),addedge(p[i],i);
    for (int i=1;i<=n;i++) if (!arr[i]) tim++,dfs(i);
    int s1=get_mex(p1);
    vis[s1]=1;
    int s2=get_mex(p1);
    vis[s1]=0;
    if (sg[p2]!=s1) {puts("POSSIBLE");return 0;}
    sg[p1]=s2;
    int x=p1;
    while (x!=p2) x=p[x],sg[x]=get_mex(x);
    if (sg[p2]==s1) puts("POSSIBLE");
    else puts("IMPOSSIBLE");
    return 0;
}
```