Problem: 

Interaction

This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.

In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.

Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.

Example

Input


Is it rated?
Is it rated?
Is it rated?


Output


NO
NO
NO

Solution (in CPP):

Solution:
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<vector>
#include<set>
#include<utility>
#include<cstdlib>
#include<map>
#include<list>
#include<deque>
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
using namespace std;
ll gcd(ll i, ll j) {
	if (j == 0)return i;
	else return gcd(j, i % j);
}
ll lcm(ll i, ll j) {
	return i * j / gcd(i, j);
}
ll max(ll i, ll j) {
	return (i > j) ? i : j;
}
ll min(ll i, ll j) {
	return (i < j) ? i : j;
}
ll numofdivs(ll n) {
	ll ans = 0;
	for (int i = 1; i <= sqrt(n); i++) {
		if (n % i == 0) {
			if (n / i == i)
				ans++;
			else
				ans += 2;
		}
	}
	return ans;
}
bool prime(ll i) {
	if (i < 2)
		return false;
	else {
		for (int j = 2; j <= i / 2; j++) {
			if (i % j == 0)
				return false;
		}
	}
	return true;
}
bool primewithexc(ll i, ll k) {
	if (i < 2)
		return false;
	else {
		for (ll j = 2; j <= i / 2; j++) {
			if (j == k)
				continue;
			if (i % j == 0)
				return false;
		}
	}
	return true;
}
bool Tprime(ll i) {
	if (sqrt(i) == (ll)sqrt(i))
		if (numofdivs(i) == 3)
			return true;
		else
			return false;
	else
		return false;
}
bool positive_negative(int n, int x)
{
	if (n < 0 && x>0)
		return true;
	else if (n > 0 && x < 0)
		return true;
	else
		return false;
}
void binary(vector<int>&v, int n) {
	int i;
	//cout << "Enter the number to convert: ";
	//cin >> n;
	for (i = 0; n > 0; i++)
	{
		v[i]= n % 2;
		n = n / 2;
	}

}
ll sumofdigs(ll n) {
	ll sum = 0;
	ll remainder = 0;
	while (n > 0) {
		remainder = n % 10;
		sum += remainder;
		n /= 10;
	}
	return sum;
}
int lcsnums(int x[], int y[], int n, int m)
{
	int lcssuf[100 + 1][100 + 1];
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= m; j++) {
			if (i == 0 || j == 0)
				lcssuf[i][j] = 0;
			else if (x[i - 1] == y[j - 1])
				lcssuf[i][j] = lcssuf[i - 1][j - 1] + 1;
			else
				lcssuf[i][j] = max(lcssuf[i - 1][j], lcssuf[i][j - 1]);
		}
	}
	return lcssuf[n][m];
}
int lcs(string x, string y, int n, int m)
{
	int lcssuf[100 + 1][100 + 1];
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= m; j++) {
			if (i == 0 || j == 0)
				lcssuf[i][j] = 0;
			else if (x[i - 1] == y[j - 1])
				lcssuf[i][j] = lcssuf[i - 1][j - 1] + 1;
			else
				lcssuf[i][j] = max(lcssuf[i - 1][j], lcssuf[i][j - 1]);
		}
	}
	return lcssuf[n][m];
}
int lcssub(string x, string y, int n, int m)
{
	int lcssuf[100+1][100+1];
	int answer = 0;
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= m; j++) {
			if (i == 0 || j == 0)
				lcssuf[i][j] = 0;
			else if (x[i - 1] == x[j - 1]) {
				lcssuf[i][j] = lcssuf[i - 1][j - 1] + 1;
				answer = max(answer, lcssuf[i][j]);
			}
			else
				lcssuf[i][j] = 0;
		}
	}
	return answer;
	//return lcssuf[n][m];

}
string mul(string s, int k) {
	string res = "";
	while (k--) res += s;
	return res;
};
int fact(int n) {
	int sum = 1;
	for (int i = 1; i <= n; i++)
		sum *= i;
	return sum;
}
void solve() {
	string x = "NO";
	string s;
	for (ll i = 0; i <=18000; i++)
		cout << x << endl;
}



int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	//int t = 1;
	//cin >> t;
	//cout << min(1, 3);
	//while (t--) {
		solve();
	//cout << ('s' - 'e');//9+13+12=34
	//}

	return 0;
}