#include using namespace std; using ll = long long; using pi = pair; using vpi = vector; using vi = vector; using vll = vector; using mpii = map; using mpll = map; using db = long double; #define pb push_back #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define lb lower_bound #define ub upper_bound const int MOD = (int)1e9 + 7; const db PI = acos((db)-1); const int dx[4]{1, 0, -1, 0}; const int dy[4]{0, 1, 0, -1}; // pretty printing template void printm(const map &mp) { cerr << "{" << endl; for (auto p : mp) { cerr << " { " << p.first << " : " << p.second << " }\n"; } cerr << "}" << endl; } template void printv(const vector &v) { cerr << "["; for (int i = 0; i < v.size(); i++) { if (i == v.size() - 1) { cerr << v[i]; } else { cerr << v[i] << ", "; } } cerr << "]\n"; } template void printvv(const vector> &v) { cerr << "[\n"; for (auto &vec : v) { cout << " "; printv(vec); } cerr << "]\n"; } void print() { cerr << "\n"; } template void print(T val, TS... vals) { cerr << val << " "; print(vals...); } /* stuff you should look for: --------------------------- * special cases (n=1?) * int overflow, array bounds * do smth instead of nothing and stay organized * WRITE STUFF DOWN * DON'T GET STUCK ON ONE APPROACH */ bool isPalindrome(string &s) { return string(all(s)) == string(rall(s)); } void solve() { int n; cin >> n; string s; cin >> s; vector pal(n); if (n & 1) pal[n / 2] = 1; for (int i = 0; i < n / 2; i++) { pal[i] = pal[n - i - 1] = s[i] == s[n - i - 1]; } pal.erase(unique(all(pal)), pal.end()); n = pal.size(); if (count(all(pal), 0) <= 2) { cout << "Yes" << '\n'; } else { cout << "No" << '\n'; } // string haystack = ""; // for (int i = 0; i < n / 2; i++) { // haystack += (pal[i] ? "R" : "W"); // } // if (regex_match(haystack, regex("WR+W"))) { // cout << "No" << '\n'; // } else { // cout << "Yes" << '\n'; // } } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int tt; cin >> tt; while (tt--) { solve(); } }