Let be a connected, directed graph with vertices numbered from to such that any vertex is reachable from vertex . In addition, any two distinct vertices, and , are connected by at most one edge .

Consider the standard DFS (Depth-First Search) algorithm starting from vertex . As every vertex is reachable, each edge of is classified by the algorithm into one of four groups:

1. tree edge: If was discovered for the first time when we traversed .
2. back edge: If was already on the stack when we tried to traverse .
3. forward edge: If was already discovered while was on the stack.
4. cross edge: Any edge that is not a tree, back, or forward edge.

To better understand this, consider the following C++ pseudocode:

// initially false
bool discovered[n]; 

// initially false
bool finished[n];   

vector<int> g[n];

void dfs(int u) {
    // u is on the stack now
    discovered[u] = true;
    for (int v: g[u]) {
        if (finished[v]) {
            // forward edge if u was on the stack when v was discovered
            // cross edge otherwise
            continue;
        }
        if (discovered[v]) {
            // back edge
            continue;
        }
        // tree edge
        dfs(v);
    }
    finished[u] = true;
    // u is no longer on the stack
}

Given four integers, , , , and , construct any graph having exactly tree edges, exactly back edges, exactly forward edges, and exactly cross edges. Then print according to the Output Format specified below.