Problem:
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Longest Common Subsequence

For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \{a,b,c,b,d,a,b\}$ and $Y = \{b,d,c,a,b,a\}$, the sequence $\{b,c,a\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\{b,c,a\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\{b,c,b,a\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\{b,c,b,a\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.

Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Input

The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.

Output

For each dataset, print the length of LCS of $X$ and $Y$ in a line.

Constraints

$1 \leq q \leq 150$

$1 \leq$ length of $X$ and $Y$ $\leq 1,000$

$q \leq 20$ if the dataset includes a sequence whose length is more than 100

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C++14 (judge verdict: Accepted)


#define REP(i, t, n)  for (int i = t; i < n; i++)
#define ALL(n)        (n).begin(),(n).end()
#include <limits.h>
#include <stdio.h>
#include <cmath>
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <map>
#include <queue>
#include <deque>
#include <functional>
#include <algorithm>
#include <cstring>
#include <unordered_map>
#include <numeric>

using namespace std;

typedef bool               BOOL;
typedef short              WORD;
typedef unsigned short     U_WORD;
typedef long               DWORD;
typedef unsigned long      U_DWORD;
typedef long long          QWORD;
typedef unsigned long long U_QWORD;

#define LEN_MAX (1000)
DWORD dpLCS[LEN_MAX + 1][LEN_MAX + 1];

void initDpTable(DWORD maxI, DWORD maxJ)
{
	REP(i, 0, maxI + 1) {
		REP(j, 0, maxJ + 1) {
			dpLCS[i][j] = 0;
		}
	}

}

void solve_main()
{
	DWORD dwNum;
	string str1, str2;
	cin >> dwNum;
	REP(loop, 0, dwNum) {
		cin >> str1;
		cin >> str2;
		initDpTable(str1.length(), str2.length());

		REP(i, 1, (str1.length() + 1)) {
			REP(j, 1, (str2.length() + 1)) {
				if (str1[i - 1] == str2[j - 1]) {
					dpLCS[i][j] = dpLCS[i - 1] [j - 1] + 1;
				} else {
					dpLCS[i][j] = max(dpLCS[i - 1][j], dpLCS[i][j - 1]);
				}
			}
		}
		cout << dpLCS[str1.length()][str2.length()] << endl;
	}
}

int main()
{
#ifdef FOR_IDE
	ifstream in("input.txt");
	cin.rdbuf(in.rdbuf());
#endif

	solve_main();
	return 0;
}