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[105][1001], b[105];
long long Abs(long long x) { return x > 0 ? x : -x; }
long long dfs(long long pos, long long statu, bool is_max, bool is_zero) {
  if (!is_max and dp[pos][statu] != -1) return dp[pos][statu];
  if (!pos) return dp[pos][statu] = 1;
  if (statu > 3) return dp[pos][statu] = 0;
  long long end = is_max ? b[pos] : 9;
  long long ans = 0;
  for (long long i = 0; i <= end; i++) {
    if (i == 0) {
      ans += dfs(pos - 1, statu, is_max && i == end, is_zero && i == 0);
    } else if (statu <= 2) {
      ans += dfs(pos - 1, statu + 1, is_max && i == end, is_zero && i == 0);
    }
  }
  if (!is_max) dp[pos][statu] = ans;
  return ans;
}
long long slove(long long x) {
  long long len = 0, y = x;
  while (x) {
    b[++len] = x % 10;
    x /= 10;
  }
  return y == 0 ? 1 : dfs(len, 0, 1, 1);
}
signed main() {
  ios::sync_with_stdio(false);
  long long a, b;
  long long test_case;
  cin >> test_case;
  memset(dp, -1, sizeof dp);
  while (test_case--) {
    cin >> a >> b;
    cout << (slove(b) - slove(a - 1));
    cout << endl;
  }
  return 0;
}