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 Python3 (judge verdict: Accepted)

import sys
readline = sys.stdin.readline
from collections import defaultdict
from bisect import bisect_left, bisect, insort_left, insort
INF = 10000
def lcs(s1, s2):
    c = defaultdict(list)
    for i, s in enumerate(s1):
        c[s].append(i)
    for s in set(s2):
        c[s].append(INF)
    dp = []
    for s2_k in s2:
        bgn_idx = 0
        for i, cur_idx in enumerate(dp):
            chr_idx = c[s2_k][bisect_left(c[s2_k], bgn_idx)]
            if chr_idx == INF:
                break
            dp[i] = min(cur_idx, chr_idx + 1)
            bgn_idx = cur_idx
        else:
            chr_idx = c[s2_k][bisect_left(c[s2_k], bgn_idx)]
            if chr_idx != INF:
                dp.append(chr_idx + 1)
    return len(dp)
def solve():
    N = int(input())
    for _ in range(N):
        x = readline().strip()
        y = readline().strip()
        print(lcs(x, y) if len(x) < len(y) else lcs(y, x))
solve()