The figure shows the tree view of directories in Windows File Explorer. When a file is selected, there is a file path shown in the above navigation bar. Now given a tree view of directories, your job is to print the file path for any selected file.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer $N$ ($\leq 10^3$), which is the total number of directories and files. Then $N$ lines follow, each gives the unique 4-digit ID of a file or a directory, starting from the unique root ID 0000. The format is that the files of depth $d$ will have their IDs indented by $d$ spaces. It is guaranteed that there is no conflict in this tree structure.

Then a positive integer $K$ ($\le 100$) is given, followed by $K$ queries of IDs.

Output Specification:

For each queried ID, print in a line the corresponding path from the root to the file in the format: 0000->ID1->ID2->...->ID. If the ID is not in the tree, print Error: ID is not found. instead.

Sample Input:

14
0000
 1234
  2234
   3234
    4234
    4235
    2333
   5234
   6234
    7234
     9999
  0001
   8234
 0002
4 9999 8234 0002 6666

Sample Output:

0000->1234->2234->6234->7234->9999
0000->1234->0001->8234
0000->0002
Error: 6666 is not found.

#include<iostream>
#include<math.h>
#include<vector>
using namespace std;

int N;//文件数量
int K;//查询的路径
struct Node {
	int depth;
	string name="";
	string father="";
	//vector child;
}nodes[10000];
vector files;//保存文件名
vector root;//保存路径

int stringtoint(string str) {
	int ans = 0;
	int len = str.size();
	for (int i = 0; i < 4; i++) {
		ans = ans + (str[len - i - 1] - '0') * pow(10, i);
	}
	return ans;
}
int main() {
	cin >> N;
	cin.ignore();
	string index;
	int ind=0;
	for (int i = 0; i < N; i++) {
		getline(cin, index);
		ind = stringtoint(index);
		nodes[ind].name = index;
		nodes[ind].depth = nodes[ind].name.size() - 4;//深度，从0开始
		files.push_back(ind);
	}
	for (int i = 1; i < N; i++) {
		for (int j = i-1; j >= 0; j--) {//从后往前找depth-1的第一个节点即此节点的父节点
			if (nodes[files[i]].depth == (nodes[files[j]].depth + 1)) {
				nodes[files[i]].father = nodes[files[j]].name;
				break;
			}
		}
	}
	//for (int i = 0; i < N; i++) {
	//	//printf("%s %04d\n", nodes[files[i]].name,nodes[files[i]].depth);
	//	cout << nodes[files[i]].name << ' ';
	//	printf("%04d\n", nodes[files[i]].depth);
	//}
	int K;
	cin >> K;
	int file;
	for (int i = 0; i < K; i++) {
		cin >> file;
		for (int i = 0; i < N; i++) {
			if (file == files[i]) {
				while (file != 0) {
					root.push_back(file);
					file = stringtoint((nodes[file].father));
				}
				root.push_back(0);
				break;
			}
		}
		if (root.size() == 0) {
			printf("Error: %04d is not found.\n", file);
		}
		else {
			//cout << root.size();
			for (int i = root.size()-1; i >=0; i--) {
				printf("%04d", root[i]);
				//cout << root[i];
				if (i >0)cout << "->";
				else cout << endl;
			}
			root.clear();
		}
	}
}