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>

using namespace std;

using ll = long long;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll MOD = 1000000007;
const ll INF = 1000000000000000000L;
#ifdef __DEBUG

/**
 * For DEBUG
 * https://github.com/ta7uw/cpp-pyprint
 */
#include "cpp-pyprint/pyprint.h"

#endif

void Main() {
    int Q;
    cin >> Q;
    while (Q--) {
        string X, Y;
        cin >> X >> Y;
        ll N = X.size();
        ll M = Y.size();
        vector<vector<ll>> dp(N + 1, vector<ll>(M + 1, 0));
        rep(i, N) {
            rep(j, M) {
                if (X[i] == Y[j]) {
                    dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + 1);
                }
                dp[i + 1][j + 1] = max({dp[i + 1][j + 1], dp[i][j + 1], dp[i + 1][j]});
            }
        }
        cout << dp[N][M] << '\n';
    }
}

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