#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 200005;

int ehead[maxn],ecnt;
struct edge {
	int u,v,next;
}edg[maxn];
void add(int u,int v) {
	edg[++ecnt]=(edge){u,v,ehead[u]};
	ehead[u]=ecnt;
}
bool vis[maxn],ins[maxn],onring[maxn],flag;
int n,a[maxn],fa[maxn],stk[maxn],top,bac[maxn];
void dfs(int u)
{
	vis[u]=ins[u]=true;
	for (int v,j=ehead[u];j&&!flag;j=edg[j].next) 
	if (ins[v=edg[j].v]) {
		onring[v]=flag=true;stk[++top]=v;
		for (int j=u;j!=v;j=fa[j])
			onring[j]=true,stk[++top]=j;
	}
	else if (!vis[v]) fa[v]=u,dfs(v);
	ins[u]=false;
}
int calc1(int u)
{
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) bac[a[v]]++;
	int w=0;while (bac[w]) ++w;
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) bac[a[v]]--;
	return w;
}
int calc2(int u)
{
	for (int j=ehead[u];j;j=edg[j].next)
		bac[a[edg[j].v]]++;
	int w=0;while (bac[w]) ++w;
	for (int j=ehead[u];j;j=edg[j].next)
		bac[a[edg[j].v]]--;
	return w;
}
void dfs2(int u)
{
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) dfs2(v);
	a[u]=calc1(u);
}
int main()
{
	#ifdef Amberframe
		freopen("arc079f.in","r",stdin);
		freopen("arc079f.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int x,i=1;i<=n;i++) scanf("%d",&x),add(x,i);
	for (int i=1;i<=n;i++) if (!vis[i]) dfs(i);
	for (int i=1;i<=n;i++) if (onring[i]) dfs2(i);
	
	int o=a[stk[1]];
	for (int i=2;i<=top;i++) a[stk[i]]=calc2(stk[i]);
	if (a[stk[top]]!=o) {puts("POSSIBLE");return 0;}
	bac[o]++;a[stk[1]]=calc1(stk[1]);bac[o]--;
	for (int i=2;i<=top;i++) a[stk[i]]=calc2(stk[i]);
	if (a[stk[top]]==o) {puts("POSSIBLE");return 0;}
	puts("IMPOSSIBLE");
	return 0;
}
### Prompt

Generate a Cpp solution 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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 200005;

int ehead[maxn],ecnt;
struct edge {
	int u,v,next;
}edg[maxn];
void add(int u,int v) {
	edg[++ecnt]=(edge){u,v,ehead[u]};
	ehead[u]=ecnt;
}
bool vis[maxn],ins[maxn],onring[maxn],flag;
int n,a[maxn],fa[maxn],stk[maxn],top,bac[maxn];
void dfs(int u)
{
	vis[u]=ins[u]=true;
	for (int v,j=ehead[u];j&&!flag;j=edg[j].next) 
	if (ins[v=edg[j].v]) {
		onring[v]=flag=true;stk[++top]=v;
		for (int j=u;j!=v;j=fa[j])
			onring[j]=true,stk[++top]=j;
	}
	else if (!vis[v]) fa[v]=u,dfs(v);
	ins[u]=false;
}
int calc1(int u)
{
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) bac[a[v]]++;
	int w=0;while (bac[w]) ++w;
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) bac[a[v]]--;
	return w;
}
int calc2(int u)
{
	for (int j=ehead[u];j;j=edg[j].next)
		bac[a[edg[j].v]]++;
	int w=0;while (bac[w]) ++w;
	for (int j=ehead[u];j;j=edg[j].next)
		bac[a[edg[j].v]]--;
	return w;
}
void dfs2(int u)
{
	for (int v,j=ehead[u];j;j=edg[j].next)
	if (!onring[v=edg[j].v]) dfs2(v);
	a[u]=calc1(u);
}
int main()
{
	#ifdef Amberframe
		freopen("arc079f.in","r",stdin);
		freopen("arc079f.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int x,i=1;i<=n;i++) scanf("%d",&x),add(x,i);
	for (int i=1;i<=n;i++) if (!vis[i]) dfs(i);
	for (int i=1;i<=n;i++) if (onring[i]) dfs2(i);
	
	int o=a[stk[1]];
	for (int i=2;i<=top;i++) a[stk[i]]=calc2(stk[i]);
	if (a[stk[top]]!=o) {puts("POSSIBLE");return 0;}
	bac[o]++;a[stk[1]]=calc1(stk[1]);bac[o]--;
	for (int i=2;i<=top;i++) a[stk[i]]=calc2(stk[i]);
	if (a[stk[top]]==o) {puts("POSSIBLE");return 0;}
	puts("IMPOSSIBLE");
	return 0;
}
```