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>
#pragma warning(disable : 4018)
using namespace std;
namespace {
vector<long long> seq;
void gen(long long x, int nz) {
  if (x > 0 && x <= (long long)1e18) {
    seq.push_back(x);
    if ((double)x * 10 < 2e18) gen(x * 10, nz);
  }
  for (int d = 1; d <= 9; d++)
    if (nz + 1 <= 3)
      if ((double)x * 10 + d < 2e18) gen(x * 10 + d, nz + 1);
}
int get(long long x) {
  if (x < 1) return 0;
  auto it = upper_bound(seq.begin(), seq.end(), x);
  if (it != seq.begin()) it--;
  return it - seq.begin() + 1;
}
void solve() {
  gen(0, 0);
  sort(seq.begin(), seq.end());
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    long long L, R;
    cin >> L >> R;
    cout << get(R) - get(L - 1) << "\n";
  }
}
}  // namespace
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(9);
  cout << fixed;
  solve();
  return 0;
}