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;
string s;
long long m;
long long dp[20][2][4];
long long dfs(long long idx, long long less, long long cnt) {
  if (cnt > 3) return 0;
  if (idx == m) return 1;
  long long& ans = dp[idx][less][cnt];
  if (ans + 1) return ans;
  ans = 0;
  long long val = less ? 9 : s[idx] - '0';
  if (!less) {
    for (long long i = 0; i < (val + 1); i++)
      ans += dfs(idx + 1, less | (i < val), cnt + (i > 0));
  } else {
    for (long long i = 0; i < (val + 1); i++)
      ans += dfs(idx + 1, less | (i < val), cnt + (i > 0));
  }
  return ans;
}
long long solve(long long n) {
  s = to_string(n);
  m = ((long long)(s).size());
  memset(dp, (long long)(-1), sizeof(dp));
  return dfs(0, 0, 0);
}
void solve() {
  long long L, R;
  cin >> L >> R;
  cout << solve(R) - solve(L - 1) << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}