Problem: Let's call some positive integer classy if its decimal representation contains no more than 3 non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000 are not.

You are given a segment [L; R]. Count the number of classy integers x such that L ≤ x ≤ R.

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^4) — the number of segments in a testcase.

Each of the next T lines contains two integers L_i and R_i (1 ≤ L_i ≤ R_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of classy integers on a segment [L_i; R_i].

Example

Input

4
1 1000
1024 1024
65536 65536
999999 1000001


Output

1000
1
0
2

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class CF1036C {

    //大数分位
    public static int[] a;

    //记忆数组
    public static long[][] dp;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for(int i = 0; i < n; i++){
            long l = sc.nextLong();
            long r = sc.nextLong();
            System.out.println(count(r) - count(l - 1));
        }
    }

    private static long count(long x){

        //数位化
        String strx = x + "";
        int len = strx.length();
        a = new int[len];
        dp = new long[len][4];

        for(int i = 0, j = strx.length(); i < j; i++){
            a[j - i - 1] = Integer.parseInt(strx.charAt(i) + "");
        }

        return dfs(len - 1,0,true);
    }

    private static long dfs(int pos, int cnt, boolean limit){
        if(pos==-1) {
            return 1;
        }

        if(!limit && dp[pos][cnt] != 0){
            return dp[pos][cnt];
        }

        int up=limit?a[pos]:9;      //当前位所需要循环的范围
        long ans = 0;               //结果
        for(int i = 0; i <= up; i++){
            if(i == 0){
                ans+=dfs(pos-1,cnt,limit && i==a[pos]);
            }else{
                if(cnt == 2){
                    ans++;
                }else{
                    ans+=dfs(pos-1,cnt+1,limit && i==a[pos]);
                }
            }
        }

        if(!limit){
            dp[pos][cnt] = ans;
        }

        return ans;
    }
}