A positive integer is called a 2-3-integer, if it is equal to 2x·3y for some non-negative integers x and y. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.

Print the number of 2-3-integers on the given segment [l, r], i. e. the number of sich 2-3-integers t that l ≤ t ≤ r.

Input

The only line contains two integers l and r (1 ≤ l ≤ r ≤ 2·109).

Output

Print a single integer the number of 2-3-integers on the segment [l, r].

Examples

Input

1 10


Output

7


Input

100 200


Output

5


Input

1 2000000000


Output

326

Note

In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.

In the second example the 2-3-integers are 108, 128, 144, 162 and 192.

import java.util.*;


import java.io.*;
public class Solution {
	static FastScanner scr=new FastScanner();
	static PrintStream out=new PrintStream(System.out);
	static StringBuilder sb=new StringBuilder();
	
	static long gcd(long a,long b){
		if(b==0) {
			return a;
		}
			
		return gcd(b,a%b);
	}
	
	
	static class pair{
		int x;int y;
		pair(int x,int y){ 
			this.x=x;this.y=y;
			
		}
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		}
	 	int[]  sort(int a[]) {
			 
	        int multiplier = 1, len = a.length, max = Integer.MIN_VALUE;
	        int b[] = new int[len];
	        int bucket[];
	        for (int i = 0; i < len; i++) if (max < a[i]) max = a[i];
	        while (max / multiplier > 0) {
	            bucket = new int[10];
	            for (int i = 0; i < len; i++) bucket[(a[i] / multiplier) % 10]++;
	            for (int i = 1; i < 10; i++) bucket[i] += (bucket[i - 1]);
	            for (int i = len - 1; i >= 0; i--) b[--bucket[(a[i] / multiplier) % 10]] = a[i];
	            for (int i = 0; i < len; i++) a[i] = b[i];
	            multiplier *= 10;
	        }
	        return a;
	    }
	 	 long modPow(long base,long exp,long mod) {
			if(exp==0) {
				return 1;
			}
			if(exp%2==0) {
				long res=(modPow(base,exp/2,mod));
				return (res*res)%mod;
			}
			return (base*modPow(base,exp-1,mod))%mod;
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	static void print(Object ele) {
		sb.append(ele).append(" ");
	}
	static void println(Object ele) {
		sb.append(ele).append('\n');
		
	}
	static void println() {
		sb.append('\n');
	}
	static long LCM(long a,long b) {
		return a*b/(gcd(a, b));
	}
//	static long n,m,k,l,q,count,sum,temp,ans;
//	static long []a;
//	static long b[];
//	static  char c[];
//	static char t[]'
	static void solve() {	
		long l=scr.nextLong();
		long r=scr.nextLong();
		long two=1;
		int count=0;
		long three=1;
		while(two<=r) {
			
			three=1;
			while(two*three<=r){
				if(two*three>=l) {
					count++;
				}
				three*=3;
			}
			two*=2;
		}
			out.println(count);
	}
	static int  MAX=Integer.MAX_VALUE;
	static int  MIN=Integer.MIN_VALUE;
	public static void main(String []args) {
//		int t=scr.nextInt();
//		while(t-->0) {
//			solve();
//		}
		solve();
		out.println(sb);
	}
}
	 