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 (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>


// Util
#define MAX(x, y) ((x) >= (y) ? (x) : (y))
//


// I/O
int get_uint() {
    int n = 0;
    int c = getchar_unlocked();
    if(c < 48 || 57 < c) return c;
    while(47 < c && c < 58) n = 10 * n + (c & 0xf), c = getchar_unlocked();
    return n;
}
int get_str(char *str) {
    int c;
    while((c = getchar_unlocked()) > 32) *str++ = (char)c;
    *str = 0;
    return c;
}
void put_uint(int n) {
    if(!n) {
        putchar_unlocked('0');
        return;
    }
    char buf[11];
    int i = 0;
    while(n) buf[i++] = (char)(n % 10 + '0'), n /= 10;
    while(i--)putchar_unlocked(buf[i]);
}
//


// Solve
int lcs(char *s, char *t) {
    const size_t M = strlen(s);
    const size_t N = strlen(t);
    int dp[1001][1001] = {};
    for(int i = 0; i < M; ++i)
        for(int j = 0; j < N; ++j)
            if(s[i] == t[j]) dp[i + 1][j + 1] = dp[i][j] + 1;
            else dp[i + 1][j + 1] = MAX(dp[i + 1][j], dp[i][j + 1]);
    return dp[M][N];
}

int main(int argc, char **argv) {
    int q = get_uint();
    char s[1001], t[1001];
    for(int i=0;i<q;++i){
        get_str(s);
        get_str(t);
        put_uint(lcs(s, t));
        putchar_unlocked('\n');
    }
    return 0;
}