/*******************************************************
Problem - Subsequences
Solution by: Sean Falconer
Algorithm: This is just a simple extension to the common
dynamic programming solution to LCS for two strings.
*******************************************************/

#include <iostream>
#include <string>
using namespace std;

int A[51][51][51];
int B[51][51][51];
string s1, s2, s3;

// print the LCS
void print(int i, int j, int k) {
	if(i == 0 || j == 0 || k == 0) return;
	if(B[i][j][k] == 1) {
		print(i-1, j-1, k-1);
		cout << s1[i-1];
	}
	else if(B[i][j][k] == 2) {
		print(i-1, j, k);
	}
	else if(B[i][j][k] == 3) {
		print(i, j-1, k);
	}
	else {
		print(i, j, k-1);
	}
}

// compute the LCS
void lcs(string & s1, string & s2, string & s3) {
	for(int i = 0; i < 51; i++) { A[i][0][0] = 0; A[0][i][0] = 0; A[0][0][i] = 0; }
	for(int i = 1; i <= s1.length(); i++) {
		for(int j = 1; j <= s2.length(); j++) {
			for(int k = 1; k <= s3.length(); k++) {
				if(s1[i-1] == s2[j-1] && s1[i-1] == s3[k-1]) {
					A[i][j][k] = A[i-1][j-1][k-1] + 1;
					B[i][j][k] = 1;
				}
				else {
					if(A[i-1][j][k] > A[i][j-1][k]) {
						A[i][j][k] = A[i-1][j][k];
						B[i][j][k] = 2;
					}
					else {
						A[i][j][k] = A[i][j-1][k];
						B[i][j][k] = 3;
					}
					if(A[i][j][k-1] > A[i][j][k]) {
						A[i][j][k] = A[i][j][k-1];
						B[i][j][k] = 4;
					}
				}
			}
		}
	}
	cout << A[s1.length()][s2.length()][s3.length()] << " ";
	print(s1.length(), s2.length(), s3.length());
	cout << endl;
}

int main() {
	int T;
	cin >> T;
	for(int i = 0; i < T; i++) {
		cin >> s1 >> s2 >> s3;
		lcs(s1, s2, s3);
	}
	return 0;
}