#include <iostream>
using namespace std;
const int N = 30;
string val[N];
int l_tr[N], r_tr[N];   //left and right collide with std::left, std::right, avoid use
bool has_father[N];     

int n;

int find_root()     //find the tree root, it don't have father node
{
    int root = 0;
    for (int i = 1; i <= n; i++ ) {
        if (!has_father[i]) root = i;
    }
    return root;
}

void dfs(int root)
{
    if (root == -1) return;
    cout << "(";                    //according to the question, each step must have brackets, begin with one, and end with another
    if (l_tr[root] != -1 and r_tr[root] != -1) {
        dfs(l_tr[root]);            //normally do postorder recursion
        dfs(r_tr[root]);
        cout << val[root];          //visit root val
    }
    else if (l_tr[root] == -1 and r_tr[root] == -1) {   //reach a leave node
        cout << val[root];          //cout node
    }
    else if (l_tr[root] == -1 and r_tr[root] != -1) {   //the condition a node '-' only has right child
        cout << val[root];          //int this condition, visit root first and dfs later
        dfs(r_tr[root]);
    }
    cout << ")";
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++ ) {
        string v; int l, r;
        cin >> v >> l >> r;
        val[i] = v;
        l_tr[i] = l, r_tr[i] = r;
        if (l != -1) has_father[l] = true;
        if (r != -1) has_father[r] = true;
    }

    int root = find_root();
    dfs(root);
    return 0;
}