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][2][4];
long long fn(long long idx, long long flag, long long cnt, string s) {
  if (cnt > 3) return 0;
  if (idx == s.size()) return 1;
  long long &ans = dp[idx][flag][cnt];
  if (ans != -1) return ans;
  ans = 0;
  if (flag == 0) {
    for (long long i = 0; i <= (s[idx] - '0'); i++) {
      long long k = 0;
      if (i == 0)
        k = 0;
      else
        k = 1;
      if (i == (s[idx] - '0')) {
        ans += fn(idx + 1, 0, cnt + k, s);
      } else {
        ans += fn(idx + 1, 1, cnt + k, s);
      }
    }
  } else {
    for (long long i = 0; i <= 9; i++) {
      long long k = 0;
      if (i == 0)
        k = 0;
      else
        k = 1;
      ans += fn(idx + 1, 1, cnt + k, s);
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    memset(dp, -1, sizeof(dp));
    cin >> x >> y;
    long long i;
    string a, b;
    x--;
    a = to_string(x);
    b = to_string(y);
    long long ans = fn(0, 0, 0, b);
    memset(dp, -1, sizeof(dp));
    long long ans1 = fn(0, 0, 0, a);
    cout << ans - ans1 << endl;
  }
}