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 mod = 1e9 + 7;
const double pi = 3.141592653589793238462643383279;
vector<long long> res;
void brute(int pos, int cnt, long long cur) {
  if (pos == 18) {
    res.push_back(cur);
    return;
  }
  brute(pos + 1, cnt, cur * 10);
  if (cnt < 3) {
    for (int i = 1; i <= 9; ++i) brute(pos + 1, cnt + 1, cur * 10 + i);
  }
}
void solve() {
  long long L, R;
  cin >> L >> R;
  cout << upper_bound((res).begin(), (res).end(), R) -
              lower_bound((res).begin(), (res).end(), L);
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long testCases = 1;
  brute(0, 0, 0);
  res.push_back(1000000000000000000);
  cin >> testCases;
  while (testCases--) {
    solve();
    cout << "\n";
  }
}