#include <bits/stdc++.h>
#define REP(i, a, b) for (int i = a; i <= b; ++i)
#define PER(i, a, b) for (int i = a; i >= b; --i)
#define RVC(i, S) for (int i = 0; i < S.size(); ++i)
#define mp make_pair
#define pb push_back
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define fi first
#define se second
using namespace std;
 
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> VI;

const int N = 200005;
int pi[N], n, out[N], on_tree[N], Q[N], sg[N], vis[N], va[N];
VI to[N];

void tok(int x){
	memset(va, 0, sizeof va);
	for (int i = pi[x]; i != x; i = pi[i]){
		RVC(j, to[i]) va[sg[to[i][j]]] = i;
		for (sg[i] = 0; va[sg[i]] == i || vis[sg[i]] == i; ++sg[i]);
	}
	// cerr << x << ' ' << sg[x] << endl;
	RVC(j, to[x]) va[sg[to[x][j]]] = x;
	int tmp = 0;
	for (; va[tmp] == x || vis[tmp] == x; ++tmp);
	if (tmp == sg[x]){
		puts("POSSIBLE");
		exit(0);
	}
}

int main(){
	scanf("%d", &n);
	REP(i, 1, n){
		scanf("%d", &pi[i]);
		to[pi[i]].pb(i);
		++out[pi[i]];
	}
	int hd = 1, tl = 0;
	REP(i, 1, n) if (!out[i]) Q[++tl] = i;
	while (hd <= tl){
		int x = Q[hd++];
		on_tree[x] = 1;
		--out[pi[x]];
		if (!out[pi[x]]) Q[++tl] = pi[x];
	}
	REP(i, 1, tl){
		int x = Q[i];
		RVC(j, to[x]) vis[sg[to[x][j]]] = x;
		for (sg[x] = 0; vis[sg[x]] == x; ++sg[x]);
	}
	REP(i, 1, n) if (!on_tree[i]){
		RVC(j, to[i]) if (on_tree[to[i][j]]) vis[sg[to[i][j]]] = i;
		for (sg[i] = 0; vis[sg[i]] == i; ++sg[i]);
		tok(i);
		for (++sg[i]; vis[sg[i]] == i; ++sg[i]);
		tok(i);
		return puts("IMPOSSIBLE"), 0;
	}
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#define REP(i, a, b) for (int i = a; i <= b; ++i)
#define PER(i, a, b) for (int i = a; i >= b; --i)
#define RVC(i, S) for (int i = 0; i < S.size(); ++i)
#define mp make_pair
#define pb push_back
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define fi first
#define se second
using namespace std;
 
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> VI;

const int N = 200005;
int pi[N], n, out[N], on_tree[N], Q[N], sg[N], vis[N], va[N];
VI to[N];

void tok(int x){
	memset(va, 0, sizeof va);
	for (int i = pi[x]; i != x; i = pi[i]){
		RVC(j, to[i]) va[sg[to[i][j]]] = i;
		for (sg[i] = 0; va[sg[i]] == i || vis[sg[i]] == i; ++sg[i]);
	}
	// cerr << x << ' ' << sg[x] << endl;
	RVC(j, to[x]) va[sg[to[x][j]]] = x;
	int tmp = 0;
	for (; va[tmp] == x || vis[tmp] == x; ++tmp);
	if (tmp == sg[x]){
		puts("POSSIBLE");
		exit(0);
	}
}

int main(){
	scanf("%d", &n);
	REP(i, 1, n){
		scanf("%d", &pi[i]);
		to[pi[i]].pb(i);
		++out[pi[i]];
	}
	int hd = 1, tl = 0;
	REP(i, 1, n) if (!out[i]) Q[++tl] = i;
	while (hd <= tl){
		int x = Q[hd++];
		on_tree[x] = 1;
		--out[pi[x]];
		if (!out[pi[x]]) Q[++tl] = pi[x];
	}
	REP(i, 1, tl){
		int x = Q[i];
		RVC(j, to[x]) vis[sg[to[x][j]]] = x;
		for (sg[x] = 0; vis[sg[x]] == x; ++sg[x]);
	}
	REP(i, 1, n) if (!on_tree[i]){
		RVC(j, to[i]) if (on_tree[to[i][j]]) vis[sg[to[i][j]]] = i;
		for (sg[i] = 0; vis[sg[i]] == i; ++sg[i]);
		tok(i);
		for (++sg[i]; vis[sg[i]] == i; ++sg[i]);
		tok(i);
		return puts("IMPOSSIBLE"), 0;
	}
	return 0;
}
```