#include <bits/stdc++.h>

const int N = 200054;

int n;
int p[N], fc[N], nc[N];
int cnt = 0, cyc[N];
int f[N], mark[N];
bool used[N];

inline void link(int x, int px) {nc[x] = fc[px], fc[px] = x;}

int dfs(int x) {
	int y, &g = f[x];
	for (y = fc[x]; y; y = nc[y]) dfs(y);
	for (y = fc[x]; y; y = nc[y]) mark[f[y]] = x;
	for (; mark[g] == x; ++g); return g;
}

int main() {
	int i, C, _;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i) scanf("%d", p + i);
	for (C = 1; !used[C]; C = p[C]) used[C] = true;
	i = C;
	do cyc[cnt++] = i, _ = p[i], p[i] = 0, i = _; while (i != C);
	if (!(cnt & 1)) return puts("POSSIBLE"), 0;
	for (i = 1; i <= n; ++i) if (p[i]) link(i, p[i]);
	dfs(*cyc);
	for (i = 1; i < cnt; ++i)
		if (dfs(cyc[i]) != dfs(*cyc)) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}
### 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>

const int N = 200054;

int n;
int p[N], fc[N], nc[N];
int cnt = 0, cyc[N];
int f[N], mark[N];
bool used[N];

inline void link(int x, int px) {nc[x] = fc[px], fc[px] = x;}

int dfs(int x) {
	int y, &g = f[x];
	for (y = fc[x]; y; y = nc[y]) dfs(y);
	for (y = fc[x]; y; y = nc[y]) mark[f[y]] = x;
	for (; mark[g] == x; ++g); return g;
}

int main() {
	int i, C, _;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i) scanf("%d", p + i);
	for (C = 1; !used[C]; C = p[C]) used[C] = true;
	i = C;
	do cyc[cnt++] = i, _ = p[i], p[i] = 0, i = _; while (i != C);
	if (!(cnt & 1)) return puts("POSSIBLE"), 0;
	for (i = 1; i <= n; ++i) if (p[i]) link(i, p[i]);
	dfs(*cyc);
	for (i = 1; i < cnt; ++i)
		if (dfs(cyc[i]) != dfs(*cyc)) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}
```