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[21][21][2];
long long solve(string &l, long long n, int cnt, long long tight) {
  if (n == 0) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (dp[n][cnt][tight] != -1) {
    return dp[n][cnt][tight];
  }
  long long ans = 0;
  long long ub = tight ? (l[l.length() - n] - '0') : 9;
  for (long long i = 0; i <= ub; i++) {
    if (cnt > 3) continue;
    if (i > 0)
      ans += solve(l, n - 1, cnt + 1, tight & (i == ub));
    else
      ans += solve(l, n - 1, cnt, tight & (i == ub));
  }
  return dp[n][cnt][tight] = ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    long long l, r, a, b;
    cin >> l >> r;
    l--;
    string L = to_string(l);
    string R = to_string(r);
    memset(dp, -1, sizeof dp);
    a = solve(L, L.length(), 0, 1);
    memset(dp, -1, sizeof dp);
    b = solve(R, R.length(), 0, 1);
    cout << b - a << endl;
  }
  return 0;
}