#include<bits/stdc++.h>
using namespace std;
/*
floor will be zero if 2^x > len
therefore 2^x <= len (i.e. n)
2^x <= n
there x <= 9
there binary representation of x can set bit only in first 4 bits
so when taking xor or two substring only last 4 bits can differ rest all
have to be same

we can brute force those last atmost 4 length substring and then maximize length
to the left of these substring as long as s[i] == t[i] .. store this max length in dp
*/
int MaxProfit(int N, string s, string t) {
      int ans = 0;
      int dp[N][N] = {};
      for(int i = 0; i < N; i++) {
            for(int j = 0; j < N; j++) {
                  if(s[i] != t[j]) continue;
                  if(i && j) dp[i][j] = 1 + dp[i-1][j-1];
            }
      }
      for(int i = 0; i < N; i++) {
            for(int j = 0; j < N; j++) {
                  int val = 0;
                  for(int k = 0; k < 4 && i+k < N && j + k < N; k++) {
                        val *= 2;
                        val += s[i+k] != t[j+k];
                        int len = k+1 + (i && j ? dp[i-1][j-1] : 0);
                        ans = max(ans, len / (1 << val));
                  }
            }
      }
      return ans;
}

int main() {
      int t;
      cin >> t;
      while(t--) {
            int n;
            string s, t;
            cin >> n >> s >> t;
            cout << MaxProfit(n, s, t) << "\n";
      }
}