// Crt. 2020-01-10  07:44:23
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define N 200005
using namespace std;

int n, fa[N], cir[N], top, inc[N], sg[N], vis[N], now[N];
int tot, fir[N], nxt[N], to[N];

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

void getcir() {
    int x = 1;
    while (!inc[x]) inc[x] = 1, x = fa[x];
    memset(inc, 0, sizeof inc);
    while (!inc[x]) cir[++top] = x, inc[x] = 1, x = fa[x];
}

void dfs(int u, int p) {
    for (int i = fir[u], v; i; i = nxt[i]) {
        v = to[i];
        if (inc[v] || p == v) continue;
        dfs(v, u);
    }
    for (int i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]] && p != to[i]) vis[sg[to[i]]] = 1;
    while (vis[sg[u]]) ++sg[u];
    for (int i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]] && p != to[i]) vis[sg[to[i]]] = 0;
}

int main(){
#ifdef LOCAL_JUDGE
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif

    cin >> n;
    for (int i = 1; i <= n; ++i)
        scanf("%d", fa + i), addE(fa[i], i);

    getcir();
    
    for (int i = 1; i <= top; ++i) dfs(cir[i], 0);

    for (int k = 0; k < 2; ++k) {
        int u; now[u = cir[1]] = sg[cir[1]] + k;

        while (fa[u] != cir[1]) {
            if (sg[fa[u]] == now[u]) now[fa[u]] = sg[fa[u]] + 1;
            else now[fa[u]] = sg[fa[u]];
            u = fa[u];
        }

        if (k == 0 && now[u] != now[fa[u]]) return puts("POSSIBLE"), 0;
        if (k == 1 && now[u] + 1 == now[fa[u]]) return puts("POSSIBLE"), 0;
    }

    puts("IMPOSSIBLE");

    return 0;
}

### 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
// Crt. 2020-01-10  07:44:23
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define N 200005
using namespace std;

int n, fa[N], cir[N], top, inc[N], sg[N], vis[N], now[N];
int tot, fir[N], nxt[N], to[N];

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

void getcir() {
    int x = 1;
    while (!inc[x]) inc[x] = 1, x = fa[x];
    memset(inc, 0, sizeof inc);
    while (!inc[x]) cir[++top] = x, inc[x] = 1, x = fa[x];
}

void dfs(int u, int p) {
    for (int i = fir[u], v; i; i = nxt[i]) {
        v = to[i];
        if (inc[v] || p == v) continue;
        dfs(v, u);
    }
    for (int i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]] && p != to[i]) vis[sg[to[i]]] = 1;
    while (vis[sg[u]]) ++sg[u];
    for (int i = fir[u]; i; i = nxt[i])
        if (!inc[to[i]] && p != to[i]) vis[sg[to[i]]] = 0;
}

int main(){
#ifdef LOCAL_JUDGE
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif

    cin >> n;
    for (int i = 1; i <= n; ++i)
        scanf("%d", fa + i), addE(fa[i], i);

    getcir();
    
    for (int i = 1; i <= top; ++i) dfs(cir[i], 0);

    for (int k = 0; k < 2; ++k) {
        int u; now[u = cir[1]] = sg[cir[1]] + k;

        while (fa[u] != cir[1]) {
            if (sg[fa[u]] == now[u]) now[fa[u]] = sg[fa[u]] + 1;
            else now[fa[u]] = sg[fa[u]];
            u = fa[u];
        }

        if (k == 0 && now[u] != now[fa[u]]) return puts("POSSIBLE"), 0;
        if (k == 1 && now[u] + 1 == now[fa[u]]) return puts("POSSIBLE"), 0;
    }

    puts("IMPOSSIBLE");

    return 0;
}

```