#include<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n, to[maxn], cnt[maxn];
std::set<int> s[maxn];
std::queue<int> q;
int sg[maxn];
inline void check(int x, int type) {
	static int cnt[maxn];
	static std::set<int> s[maxn];
	rep(i, 1, n) cnt[i] = ::cnt[i], s[i] = ::s[i];
	q.push(to[x]), s[to[x]].insert(sg[x]), cnt[x] = 0;
	for(;!q.empty();) {
		int x = q.front(); q.pop(); sg[x] = 0;
		for(;s[x].count(sg[x]);) ++ sg[x];
		s[to[x]].insert(sg[x]);
		if(--cnt[to[x]] == 0) q.push(to[x]);
	}
	if(type ? !s[x].count(sg[x]) : s[x].count(sg[x] - 1)) {
		cout << "POSSIBLE" << '\n';
		exit(0);
	}
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n) {
		cin >> to[i];
		++ cnt[to[i]];
	}
	rep(i, 1, n) if(!cnt[i]) {
		q.push(i);
	}
	for(;!q.empty();) {
		int x = q.front(); q.pop();
		for(;s[x].count(sg[x]);) ++ sg[x];
		s[to[x]].insert(sg[x]);
		if(--cnt[to[x]] == 0) q.push(to[x]);
	}
	rep(i, 1, n) if(cnt[i]) {
		for(;s[i].count(sg[i]);) ++ sg[i];
		check(i, 1);
		++ sg[i];
		check(i, 0);
		break;
	}
	cout << "IMPOSSIBLE" << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n, to[maxn], cnt[maxn];
std::set<int> s[maxn];
std::queue<int> q;
int sg[maxn];
inline void check(int x, int type) {
	static int cnt[maxn];
	static std::set<int> s[maxn];
	rep(i, 1, n) cnt[i] = ::cnt[i], s[i] = ::s[i];
	q.push(to[x]), s[to[x]].insert(sg[x]), cnt[x] = 0;
	for(;!q.empty();) {
		int x = q.front(); q.pop(); sg[x] = 0;
		for(;s[x].count(sg[x]);) ++ sg[x];
		s[to[x]].insert(sg[x]);
		if(--cnt[to[x]] == 0) q.push(to[x]);
	}
	if(type ? !s[x].count(sg[x]) : s[x].count(sg[x] - 1)) {
		cout << "POSSIBLE" << '\n';
		exit(0);
	}
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n) {
		cin >> to[i];
		++ cnt[to[i]];
	}
	rep(i, 1, n) if(!cnt[i]) {
		q.push(i);
	}
	for(;!q.empty();) {
		int x = q.front(); q.pop();
		for(;s[x].count(sg[x]);) ++ sg[x];
		s[to[x]].insert(sg[x]);
		if(--cnt[to[x]] == 0) q.push(to[x]);
	}
	rep(i, 1, n) if(cnt[i]) {
		for(;s[i].count(sg[i]);) ++ sg[i];
		check(i, 1);
		++ sg[i];
		check(i, 0);
		break;
	}
	cout << "IMPOSSIBLE" << '\n';
}

```