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;
const int dsz = 18;
const long long inf = 1e18;
int ts;
long long pw[dsz + 5];
vector<long long> dat;
void inline init() {
  pw[0] = 1;
  for (int i = 1; i <= dsz; ++i) {
    pw[i] = pw[i - 1] * 10;
  }
  unsigned long long cur;
  for (int d1 = 0; d1 <= 9; ++d1)
    for (int d2 = 0; d2 <= 9; ++d2)
      for (int d3 = 0; d3 <= 9; ++d3) {
        for (int i1 = 0; i1 <= 16; ++i1)
          for (int i2 = i1 + 1; i2 <= 17; ++i2)
            for (int i3 = i2 + 1; i3 <= 18; ++i3) {
              cur = pw[i1] * d1 + pw[i2] * d2 + pw[i3] * d3;
              if (cur > inf) break;
              dat.push_back(cur);
            }
      }
  sort(dat.begin(), dat.end());
  dat.resize(unique(dat.begin(), dat.end()) - dat.begin());
}
int main() {
  init();
  cin >> ts;
  for (int cs = 1; cs <= ts; ++cs) {
    long long l, r;
    cin >> l >> r;
    cout << upper_bound(dat.begin(), dat.end(), r) -
                lower_bound(dat.begin(), dat.end(), l)
         << '\n';
  }
}