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;
long long mod = 1000000007;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void gen(int ind, int cnt, long long num, vector<long long>& ans) {
  if (ind == 18) {
    ans.push_back(num);
    return;
  }
  gen(ind + 1, cnt, num * 10, ans);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      gen(ind + 1, cnt + 1, num * 10 + i, ans);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  long long t;
  cin >> t;
  vector<long long> ans;
  gen(0, 0, 0, ans);
  sort(ans.begin(), ans.end());
  ans.push_back(1000000000000000000);
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << upper_bound(ans.begin(), ans.end(), r) -
                upper_bound(ans.begin(), ans.end(), (l - 1))
         << endl;
  }
  return 0;
}