#include "bits/stdc++.h"
using namespace std;
#define maxn 200001
#define rep(i,n) for(int i=0; i<int(n); ++i)
#define pb push_back
#define mp make_pair
#define mod 1000000007

int n, m, x, y;
int a[maxn], b[maxn], back[maxn], nxt[maxn], ans[maxn];
bool vis[maxn];
vector<int> c, e[maxn];
void circle(){
	int hd;
	for (int u = 1;; u = back[u]){
		if (vis[u]){
			hd = u;
			break;
		}
		vis[u] = 1;
	}
	for (int u = hd; ;){
		c.pb(u);
		nxt[back[u]] = u; 
		u = back[u];
		if (u == hd) break;
	}
}
void dfs(int u){
	vector<int> tmp;
	tmp.clear();
	for (int v : e[u]){
		if (v != nxt[u]){
			dfs(v);
			tmp.pb(a[v]);
		}
	}
	sort(tmp.begin(), tmp.end());
	tmp.erase(unique(tmp.begin(), tmp.end()), tmp.end());
	int i = 0;
	while (tmp.size() > i && tmp[i] == i) ++i;
	a[u] = i;
	while (tmp.size() > i && tmp[i] == i + 1) ++i;
	b[u] = i + 1;
	//cout << u << ' ' << a[u] << ' ' << b[u] << endl;
}
void solve(){
	cin >> n;
	for (int i = 1; i <= n; ++i){
		cin >> x;
		e[x].pb(i);
		back[i] = x;
	}
	circle();
	for (int i : c) dfs(i);
	//
	ans[c[0]] = a[c[0]];
	for (int u = back[c[0]]; u != c[0]; u = back[u]){
		ans[u] = (ans[nxt[u]] == a[u]) ? b[u] : a[u];
	}
	if (a[c[0]] != ans[nxt[c[0]]]){
		puts("POSSIBLE");
		return;
	}
	ans[c[0]] = b[c[0]];
	for (int u = back[c[0]]; u != c[0]; u = back[u]){
		ans[u] = (ans[nxt[u]] == a[u]) ? b[u] : a[u];
	}
	if (a[c[0]] == ans[nxt[c[0]]]){
		puts("POSSIBLE");
		return;
	}
	puts("IMPOSSIBLE");
}

int main(int argc, char const *argv[])
{
    ios_base::sync_with_stdio(0);
    solve();
}
### 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 "bits/stdc++.h"
using namespace std;
#define maxn 200001
#define rep(i,n) for(int i=0; i<int(n); ++i)
#define pb push_back
#define mp make_pair
#define mod 1000000007

int n, m, x, y;
int a[maxn], b[maxn], back[maxn], nxt[maxn], ans[maxn];
bool vis[maxn];
vector<int> c, e[maxn];
void circle(){
	int hd;
	for (int u = 1;; u = back[u]){
		if (vis[u]){
			hd = u;
			break;
		}
		vis[u] = 1;
	}
	for (int u = hd; ;){
		c.pb(u);
		nxt[back[u]] = u; 
		u = back[u];
		if (u == hd) break;
	}
}
void dfs(int u){
	vector<int> tmp;
	tmp.clear();
	for (int v : e[u]){
		if (v != nxt[u]){
			dfs(v);
			tmp.pb(a[v]);
		}
	}
	sort(tmp.begin(), tmp.end());
	tmp.erase(unique(tmp.begin(), tmp.end()), tmp.end());
	int i = 0;
	while (tmp.size() > i && tmp[i] == i) ++i;
	a[u] = i;
	while (tmp.size() > i && tmp[i] == i + 1) ++i;
	b[u] = i + 1;
	//cout << u << ' ' << a[u] << ' ' << b[u] << endl;
}
void solve(){
	cin >> n;
	for (int i = 1; i <= n; ++i){
		cin >> x;
		e[x].pb(i);
		back[i] = x;
	}
	circle();
	for (int i : c) dfs(i);
	//
	ans[c[0]] = a[c[0]];
	for (int u = back[c[0]]; u != c[0]; u = back[u]){
		ans[u] = (ans[nxt[u]] == a[u]) ? b[u] : a[u];
	}
	if (a[c[0]] != ans[nxt[c[0]]]){
		puts("POSSIBLE");
		return;
	}
	ans[c[0]] = b[c[0]];
	for (int u = back[c[0]]; u != c[0]; u = back[u]){
		ans[u] = (ans[nxt[u]] == a[u]) ? b[u] : a[u];
	}
	if (a[c[0]] == ans[nxt[c[0]]]){
		puts("POSSIBLE");
		return;
	}
	puts("IMPOSSIBLE");
}

int main(int argc, char const *argv[])
{
    ios_base::sync_with_stdio(0);
    solve();
}
```