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 L, R;
string s;
int len;
long long memo[25][4][2];
long long dfs(int cur, int M, int f) {
  if (cur == len) return 1;
  if (memo[cur][M][f] != -1) return memo[cur][M][f];
  int v;
  if (f)
    v = s[cur] - '0';
  else
    v = 9;
  long long ans = 0;
  for (int i = 0; i <= v; i++) {
    if (i == 0)
      ans += dfs(cur + 1, M, f & (i == v));
    else {
      if (M == 3) continue;
      ans += dfs(cur + 1, M + 1, f & (i == v));
    }
  }
  return memo[cur][M][f] = ans;
}
long long solve(string& x) {
  s = x;
  len = s.length();
  memset(memo, -1, sizeof(memo));
  return dfs(0, 0, 1);
}
int chk(string& x) {
  int cnt = 0;
  for (int i = 0; i < x.length(); i++) {
    if (x[i] != '0') cnt++;
  }
  return (cnt <= 3);
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> L >> R;
    cout << solve(R) - solve(L) + chk(L) << endl;
  }
  return 0;
}