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)

#include <bits/stdc++.h>
 
#define rep(i,n) for(ll i=0;i<n;++i)
#define ALL(a) (a).begin(),(a).end()
 
using namespace std;
using ll  = long long;
using pll = pair<ll, ll>;
 
const ll MOD  = 1e9+7;
const ll LINF = 1LL << 60;
const int INF = 1e9 + 7;

int solve(string x, string y){
    //dp[i][j] xの i文字目、yのj文字目までの最長共通部分列の長さ
    int dp[1010][1010] = {};

    rep(i, x.size())rep(j, y.size()){
        dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
        if(x[i] == y[j])dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]+1);
    }

    return dp[x.size()][y.size()];
}
int main(){
    int q;
    cin >> q;

    int ans[q];

    rep(i, q){
        string x, y;
        cin >> x >> y;
        
        ans[i] = solve(x, y);
    }

    rep(i, q)cout << ans[i] << endl;
}