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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int T;
long long L[10000], R[10000];
long long solve(string A) {
  int L = A.length();
  long long dp[L + 1][4][2];
  for (int l = 0; l < (int)(L + 1); l++)
    for (int i = 0; i < (int)(4); i++)
      for (int isless = 0; isless < (int)(2); isless++) dp[l][i][isless] = 0;
  dp[0][0][0] = 1;
  for (int l = 0; l < (int)(L); l++)
    for (int i = 0; i < (int)(4); i++)
      for (int isless = 0; isless < (int)(2); isless++) {
        int a = A[l] - '0';
        for (int num = 0; num < (int)(10); num++) {
          if (num < a) {
            if (num == 0) {
              dp[l + 1][i][1] += dp[l][i][isless];
            } else if (i < 3) {
              dp[l + 1][i + 1][1] += dp[l][i][isless];
            }
          } else if (num == a || isless) {
            if (num == 0) {
              dp[l + 1][i][isless] += dp[l][i][isless];
            } else if (i < 3) {
              dp[l + 1][i + 1][isless] += dp[l][i][isless];
            }
          }
        }
      }
  long long ans = 0;
  for (int i = 0; i < (int)(4); i++) ans += dp[L][i][0] + dp[L][i][1];
  return ans;
}
int main() {
  cin >> T;
  for (int i = 0; i < (int)(T); i++) cin >> L[i] >> R[i];
  for (int t = 0; t < (int)(T); t++) {
    long long a1 = solve(to_string(L[t] - 1));
    long long a2 = solve(to_string(R[t]));
    cout << a2 - a1 << endl;
  }
}