#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#define N 500005
#define rint register int
using namespace std;

int tot = 0, fir[N], nxt[N << 1], to[N << 1];

inline void addEdge(int u, int v) {
    nxt[++tot] = fir[u], fir[u] = tot, to[tot] = v;
}

int n, inc[N], cntc, cir[N], val[N], use[N], pre[N];

inline void getcircle() {
    int x = 1;
    while (!inc[x]) inc[x] = 1, x = pre[x];
    memset(inc, 0, sizeof inc);
    while (!inc[x]) inc[cir[++cntc] = x] = 1, x = pre[x];
}

void getmex(int u) {
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            getmex(to[i]);
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            use[val[to[i]]] = 1;
    val[u] = 0;
    while (use[val[u]]) ++val[u];
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            use[val[to[i]]] = 0;
}

int main() {
    cin >> n;
    for (rint i = 1, u; i <= n; ++i)
        scanf("%d", pre + i), addEdge(pre[i], i);
    getcircle();
    int mx = 0, mn = n + 1;
    for (rint i = 1; i <= n; ++i) {
        if (!inc[i]) continue;
        getmex(i);
        mx = max(mx, val[i]);
        mn = min(mn, val[i]);
    }
    if (cntc % 2 && mx == mn) puts("IMPOSSIBLE");
    else puts("POSSIBLE");
    return 0;
}

### Prompt

In cpp, your task is to solve 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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#define N 500005
#define rint register int
using namespace std;

int tot = 0, fir[N], nxt[N << 1], to[N << 1];

inline void addEdge(int u, int v) {
    nxt[++tot] = fir[u], fir[u] = tot, to[tot] = v;
}

int n, inc[N], cntc, cir[N], val[N], use[N], pre[N];

inline void getcircle() {
    int x = 1;
    while (!inc[x]) inc[x] = 1, x = pre[x];
    memset(inc, 0, sizeof inc);
    while (!inc[x]) inc[cir[++cntc] = x] = 1, x = pre[x];
}

void getmex(int u) {
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            getmex(to[i]);
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            use[val[to[i]]] = 1;
    val[u] = 0;
    while (use[val[u]]) ++val[u];
    for (rint i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]])
            use[val[to[i]]] = 0;
}

int main() {
    cin >> n;
    for (rint i = 1, u; i <= n; ++i)
        scanf("%d", pre + i), addEdge(pre[i], i);
    getcircle();
    int mx = 0, mn = n + 1;
    for (rint i = 1; i <= n; ++i) {
        if (!inc[i]) continue;
        getmex(i);
        mx = max(mx, val[i]);
        mn = min(mn, val[i]);
    }
    if (cntc % 2 && mx == mn) puts("IMPOSSIBLE");
    else puts("POSSIBLE");
    return 0;
}

```