Problem: Recently Petya walked in the forest and found a magic stick.

Since Petya really likes numbers, the first thing he learned was spells for changing numbers. So far, he knows only two spells that can be applied to a positive integer: 

  1. If the chosen number a is even, then the spell will turn it into 3a/2; 
  2. If the chosen number a is greater than one, then the spell will turn it into a-1. 



Note that if the number is even and greater than one, then Petya can choose which spell to apply.

Petya now has only one number x. He wants to know if his favorite number y can be obtained from x using the spells he knows. The spells can be used any number of times in any order. It is not required to use spells, Petya can leave x as it is.

Input

The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^9) — the current number and the number that Petya wants to get.

Output

For the i-th test case print the answer on it — YES if Petya can get the number y from the number x using known spells, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


7
2 3
1 1
3 6
6 8
1 2
4 1
31235 6578234


Output


YES
YES
NO
YES
NO
YES
YES

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;

public class problem2 {
	public static HashSet<Long> hSet = new HashSet<>();
	public static void divisior(long n) {
		while(n % 2 == 0) {
			hSet.add((long)2);
			n /= 2;
		}
		for(long i = 3; i * i <= n; i++) {
			while(n % i == 0) {
				n /= i;
				hSet.add((long)i);
			}
		}
		if(n > 2)hSet.add(n);
	}
	public static long gcd(long a,long b) {
		if(b == 0)return(a);
		else return(gcd(b,a%b));
	}
	public static void merge(int arr[], int l, int m, int r) 
	    { 
	        // Find sizes of two subarrays to be merged 
	        int n1 = m - l + 1; 
	        int n2 = r - m; 
	  
	        /* Create temp arrays */
	        int L[] = new int [n1]; 
	        int R[] = new int [n2]; 
	  
	        /*Copy data to temp arrays*/
	        for (int i=0; i<n1; ++i) 
	            L[i] = arr[l + i]; 
	        for (int j=0; j<n2; ++j) 
	            R[j] = arr[m + 1+ j]; 
	  
	  
	        /* Merge the temp arrays */
	  
	        // Initial indexes of first and second subarrays 
	        int i = 0, j = 0; 
	  
	        // Initial index of merged subarry array 
	        int k = l; 
	        while (i < n1 && j < n2) 
	        { 
	            if (L[i] <= R[j]) 
	            { 
	                arr[k] = L[i]; 
	                i++; 
	            } 
	            else
	            { 
	                arr[k] = R[j]; 
	                j++; 
	            } 
	            k++; 
	        } 
	  
	        /* Copy remaining elements of L[] if any */
	        while (i < n1) 
	        { 
	            arr[k] = L[i]; 
	            i++; 
	            k++; 
	        } 
	  
	        /* Copy remaining elements of R[] if any */
	        while (j < n2) 
	        { 
	            arr[k] = R[j]; 
	            j++; 
	            k++; 
	        } 
	    } 
	  
	    // Main function that sorts arr[l..r] using 
	    // merge() 
	  public static void sort(int arr[], int l, int r) 
	    { 
	        if (l < r) 
	        { 
	            // Find the middle point 
	            int m = (l+r)/2; 
	  
	            // Sort first and second halves 
	            sort(arr, l, m); 
	            sort(arr , m+1, r); 
	  
	            // Merge the sorted halves 
	            merge(arr, l, m, r); 
	        } 
	    } 
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		for(int y=0;y<t;y++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			int m = Integer.parseInt(st.nextToken());
			if(m<=n)System.out.println("YES");
			else if(n == 1 && m != 1)System.out.println("NO");
			else if(n == 2 && m>3 || n== 3 && m >3)System.out.println("NO");
			else System.out.println("YES");
		}
	}
}