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:

#include <bits/stdc++.h>
#include <cstdlib> 
using namespace std;
typedef long long ll; 
bool isPrime(long long int n) 
{ 
    // Corner cases 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
  
    // This is checked so that we can skip 
    // middle five numbers in below loop 
    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
  
    for (long long int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
  
    return true; 
}
unsigned int countBits(unsigned int n) 
{ 
unsigned int count = 0; 
while (n) 
{ 
        count++; 
        n >>= 1; 
    } 
    return count; 
} 
bool isPerfectSquare(long double x) 
{   
  // Find floating point value of  
  // square root of x. 
  long double sr = sqrt(x); 
  
  // If square root is an integer 
  return ((sr - floor(sr)) == 0); 
}
int subArraySum(int arr[], int n) 
{ 
    /* Initialize curr_sum as value of  
    first element and starting point as 0 */
    int curr_sum = arr[0], start = 0, i,count=0,len; 
  len=0;
    /* Add elements one by one to curr_sum and  
    if the curr_sum exceeds the sum, 
    then remove starting element */
    for (i = 1; i <= n; i++) { 
        // If curr_sum exceeds the sum, 
        // then remove the starting elements 
        len++;
        while (curr_sum > len && start < i - 1) { 
          
            curr_sum = curr_sum - arr[start]; 
            start++; 
            len++;
        } 
  
        // If curr_sum becomes equal to sum, 
        // then return true 
        if (curr_sum == len) { 
            count++;
            
        } 
  
        // Add this element to curr_sum 
        if (i < n && i>0) {
            curr_sum = curr_sum + arr[i]; 
           
        }

    } 
  
    // If we reach here, then no subarray 
    return count;
} 

 void printDigit(int N, vector<long long int> &arr) 
{ 
    // To store the digit 
    // of the number N 
    
    int i = 0; 
    int j, r; 
  
    // Till N becomes 0 
    while (N != 0) { 
  
        // Extract the last digit of N 
        r = N % 10; 
  
        // Put the digit in arr[] 
        arr.push_back(r); 
        i++; 
  
        // Update N to N/10 to extract 
        // next last digit 
        N = N / 10; 
    } 
  
    // Print the digit of N by traversing 
    // arr[] reverse 
    // for (j = i - 1; j > -1; j--) { 
    //     printf("%d ", arr[j]); 
    // } 
}
// void solver1(vector<long long int> arr, long long int start, long long int& count, long long int a1){
//     long long int sum=0;
// for(long long int i=start+1;i<arr.size();i++){
//         sum+=arr. Find the minimum integer [i];
//         if(sum==a1 && i==arr.size()-1)
//         count++;
//     }
// }
// void solver(vector<long long int> arr, long long int start, long long int& count, long long int a1){
//     long long int sum=0;
// for(long long int i=start+1;i<arr.size()-1;i++){
//         sum+=arr[i];
//         if(sum==a1)
//         count++;
//     }
// }
bool checkDivisibility(long long int n, int digit) 
{ 
    // If the digit divides the number 
    // then return true else return false. 
    if(digit==0)
    return true;
    return (digit != 0 && n % digit == 0); 
} 
  
// Function to check if all digits 
// of n divide it or not 
bool allDigitsDivide( long long int n) 
{ 
  long long int temp = n; 
    while (temp > 0) { 
  
        // Taking the digit of the 
        // number into digit var. 
        int digit = temp % 10; 
        if (!(checkDivisibility(n, digit))) 
            return false; 
  
        temp /= 10; 
    } 
    return true; 
}
bool isPrime(int n)  
{  
    // Corner cases  
    if (n <= 1)  return false;  
    if (n <= 3)  return true;  
    
    // This is checked so that we can skip   
    // middle five numbers in below loop  
    if (n%2 == 0 || n%3 == 0) return false;  
    
    for (int i=5; i*i<=n; i=i+6)  
        if (n%i == 0 || n%(i+2) == 0)  
           return false;  
    
    return true;  
}  
  bool isPowerOfTwo (unsigned long long x) 
{ 
    /* First x in the below expression is for the case when x is 0 */
    return x && (!(x&(x-1))); 
} 
// Function to return the smallest 
// prime number greater than N 
int nextPrime(int N) 
{ 
  
    // Base case 
    if (N <= 1) 
        return 2; 
  
    int prime = N; 
    bool found = false; 
  
    // Loop continuously until isPrime returns 
    // true for a number greater than n 
    while (!found) { 
        prime++; 
  
        if (isPrime(prime)) 
            found = true; 
    } 
  
    return prime; 
} 
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
 string s;
 cin>>s;
 while(getline(cin,s)){
     if(s.length()==0)
     break;
     else
     cout<<"NO"<<endl;
 }
   //}
return 0;
 
//  0 1 2 4 5 6 7
//  1 3 1 2 1 1 1

}