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 memo[20][2][4];
vector<int> s;
long long dp(int pos, bool menor, int cnt) {
  if (cnt > 3) return 0LL;
  if (pos == s.size()) return 1LL;
  if (memo[pos][menor][cnt] != -1) return memo[pos][menor][cnt];
  long long &ans = memo[pos][menor][cnt] = 0;
  if (menor) {
    for (int i = 0; i < 10; i++) {
      ans += dp(pos + 1, menor, cnt + (i != 0));
    }
  } else {
    for (int i = 0; i <= (s[pos]); i++) {
      ans += dp(pos + 1, (i < s[pos]), cnt + (i != 0));
    }
  }
  return ans;
}
vector<int> f(long long x) {
  vector<int> ans;
  while (x) {
    ans.push_back((x % 10));
    x /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    a--;
    s = f(b);
    memset(memo, -1, sizeof memo);
    long long ans = dp(0, 0, 0) - 1;
    if (a) {
      s = f(a);
      memset(memo, -1, sizeof memo);
      ans -= (dp(0, 0, 0) - 1);
    }
    cout << ans << '\n';
  }
  return 0;
}