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 dp[20][5][2];
string c;
long long solve(long long pos, int cnt, bool flag) {
  if (pos > c.size()) return (cnt <= 3);
  if (cnt > 3) return 0;
  if (dp[pos][cnt][flag] != -1) return dp[pos][cnt][flag];
  int limit = 9;
  if (!flag) limit = c[pos - 1] - '0';
  long long res = 0;
  for (long long i = 0; i <= limit; i++) {
    if (flag || i < limit)
      res += solve(pos + 1, cnt + (i != 0 ? 1 : 0), true);
    else
      res += solve(pos + 1, cnt + (i != 0 ? 1 : 0), false);
  }
  return dp[pos][cnt][flag] = res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    l--;
    c = to_string(l);
    memset(dp, -1, sizeof(dp));
    long long resa = solve(1, 0, false);
    memset(dp, -1, sizeof(dp));
    c = to_string(r);
    long long resb = solve(1, 0, false);
    cout << resb - resa << "\n";
  }
  return 0;
}