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.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    for(int i = 0; i < t; i++) {
      long L = sc.nextLong();
      long R = sc.nextLong();
      System.out.println(func(R) - func(L - 1));
    }
  }

  public static long func(long n) {
    String str = String.valueOf(n);
    int len = str.length();
    long[][][] dp = new long[len][4][2];
    int first = Integer.parseInt(String.valueOf(str.charAt(0)));
    dp[0][0][1] = 1;
    dp[0][1][0] = 1;
    dp[0][1][1] = first - 1;

    for(int i = 1; i < len; i++) {
      int d = Integer.parseInt(String.valueOf(str.charAt(i)));
      if(d == 0) {
        for(int j = 0; j < 4; j++) {
          dp[i][j][0] = dp[i - 1][j][0];
        }
        for(int j = 0; j < 4; j++) {
          if(j >= 1) {
            dp[i][j][1] = 9 * dp[i - 1][j - 1][1] + dp[i - 1][j][1];
          } else {
            dp[i][j][1] = dp[i - 1][j][1];
          }
        }
      } else {
        for(int j = 1; j < 4; j++) {
          dp[i][j][0] = dp[i - 1][j - 1][0];
        }
        for(int j = 0; j < 4; j++) {
          if(j >= 1) {
            dp[i][j][1] = 9 * dp[i - 1][j - 1][1] + dp[i - 1][j][1] + (long)(d - 1) * dp[i - 1][j - 1][0] + dp[i - 1][j][0];
          } else {
            dp[i][j][1] = dp[i - 1][j][1] + dp[i - 1][j][0];
          }
        }
      }
    }

    long ans = 0;
    for(int j = 0; j < 4; j++) {
      for(int k = 0; k < 2; k++) {
        ans += dp[len - 1][j][k];
      }
    }

    return ans;
  }
}