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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<deque>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#include<functional>
#include<unordered_set>
#include<iomanip>
#include<cassert>
#include<regex>
#include<bitset>
#include<complex>
#include<chrono>
#include<random>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define rrep(i,a) for(int i=(int)a-1;i>=0;--i)
#define REP(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define RREP(i,a,b) for(int i=(int)a-1;i>=b;--i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
using ll=long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;

ll gcd(ll n, ll m) {
	ll tmp;
	while (m!=0) {
		tmp = n % m;
		n = m;
		m = tmp;
	}
	return n;
}

ll lcm(ll n, ll m) {
	return abs(n) / gcd(n, m)*abs(m);//gl=xy
}
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
using namespace std;

int LCS(string s,string t){
  int n=s.size(),m=t.size();
  vector<vector<int>>dp(n+1,vector<int>(m+1));
  for(int i=0;i<n;++i){
    for(int j=0;j<m;++j){
      if(s[i]==t[j])dp[i+1][j+1]=dp[i][j]+1;
      else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
    }
  }
  return dp[n][m];
}


void solve(){
  int t;
  cin>>t;
  while(t--){
    string s,t;
    cin>>s>>t;
    cout<<LCS(s,t)<<"\n";
  }
}

int main(){
	ios::sync_with_stdio(false);
  cin.tie(0);
	cout<<fixed<<setprecision(15);
  solve();
	return 0;
}