[contest:https://mirror.codeforces.com/contest/2178]

Key Observation: ==================

1.The "Yes" is Persistent: In the rules, Y OR N = Y. This means a Y never disappears; it just "**eats**" the N next to it. 2.The Trap: If you have two Ys, they will eventually have to meet as the string gets shorter. Because a Y cannot be turned back into an N, those two Ys will eventually be forced to merge. 3.The Rule: If the string has 0 or 1 Y, you can always merge all N's together first, then merge that single N with the single Y. But if there are greater than equal (>=) 2 Ys, you will eventually be forced to break the rule.

Implementation: ==================

include <bits/stdc++.h>

using namespace std;

int main() { ios::sync_with_stdio(false); cin.tie(NULL); int test; cin >> test; while (test--) { string s; cin >> s; int Ycount = 0; for (char c : s) { if (c == 'Y') Ycount++; }

if (Ycount > 1) cout << "NO" << "\n";

    else cout << "YES" << "\n"; 
}

return 0;

}