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>
const long long INF = 1e18;
long long pw[18];
std::vector<long long> v;
void init() {
  pw[0] = 1;
  for (int i = 1; i < 18; ++i) {
    pw[i] = pw[i - 1] * 10;
  }
  for (int i = 0; i < 18; ++i) {
    for (int x = 1; x < 10; ++x) {
      v.push_back(x * pw[i]);
      for (int j = 0; j < i; ++j) {
        for (int y = 1; y < 10; ++y) {
          v.push_back(x * pw[i] + y * pw[j]);
          for (int k = 0; k < j; ++k) {
            for (int z = 1; z < 10; ++z) {
              v.push_back(x * pw[i] + y * pw[j] + z * pw[k]);
            }
          }
        }
      }
    }
  }
  std::sort(v.begin(), v.end());
  v.erase(std::unique(v.begin(), v.end()), v.end());
}
int solve(long long x) {
  return std::upper_bound(v.begin(), v.end(), x) - v.begin() + (x == INF);
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  init();
  int T = 1;
  std::cin >> T;
  while (T--) {
    long long l, r;
    std::cin >> l >> r;
    std::cout << solve(r) - solve(l - 1) << "\n";
  }
}