可以观察到 Bob 操作的奇偶性，如果 Bob 操作奇数次，那么两个字符串使用逆序比较；如果 Bob 操作偶数次，那么两个字符串使用正序比较。Alice 和 Bob 轮流操作，Alice 先操作，只需要统计正序比较和逆序比较两个字符串不同字符的个数，确定 Alice 的操作次数，然后再根据奇偶性确定 Bob 操作次数。

// Date: Sat Dec 16 08:36:02 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

int tc, n;
string s, t;

int main(void) {
#ifdef _DEBUG
  freopen("1834c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> tc;

  while (tc--) {
    cin >> n >> s >> t;

    int cnt = 0, cnt1 = 0;
    For(i, 0, n) {
      if (s[i] != t[i])
        cnt++;
    }
    for (int i = 0, j = n - 1; i < n; i++, j--) {
      if (s[i] != t[j])
        cnt1++;
    }

    int res = INF;
    if (cnt & 1) {
      if (cnt > 0)
        res = min(res, cnt - 1 + cnt);
    } else {
      res = min(res, cnt + cnt);
    }

    if (cnt1 & 1) {
      res = min(res, cnt1 + cnt1);
    } else {
      if (cnt1 > 0)
        res = min(res, cnt1 - 1 + cnt1);
      else
        res = min(res, 2);
    }

    cout << res << '\n';
  }

  return 0;
}