// Problem: A1. Prefix Flip (Easy Version)
// Contest: Codeforces - Codeforces Round #658 (Div. 1)
// URL: https://codeforces.com/problemset/problem/1381/A1
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-02-27 17:49:17
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
vector<int> ans;
void solve() {
ans.clear();
int n;
cin >> n;
string a, b;
cin >> a >> b;
int i = n - 1, len = 0;

for (int j = n - 1; j >= 0; j --) {
int i = j;
while (i >= 0 && a[i] == b[i]) {
i --;
}
if (i >= 0) {
ans.pb(i + 1);
ans.pb(1);
ans.pb(i + 1);
}
j = i;
}

cout << ans.size() << " ";
for (auto t : ans) cout << t << " ";
puts("");
}
int main () {
int t;
t =1;
cin >> t;
while (t --) solve();
return 0;
}


由题目最多3*n次,所以每一次我们可以只恢复一个。这样就能恢复全部了。而每次只需要3次,所以刚好到3*n次。不会超过限制 这次是通过玩的获得思路的。所以玩是和重要的