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 s;
long long dp[20][20][2];
long long calc(int pos, int cnt_non_zero, int is) {
  if (pos == s.size()) {
    if (cnt_non_zero <= 3) return 1;
    return 0;
  }
  if (dp[pos][cnt_non_zero][is] != -1) return dp[pos][cnt_non_zero][is];
  int limit = is ? 9 : s[pos] - '0';
  long long res = 0;
  for (int digit = 0; digit <= limit; ++digit) {
    res += calc(pos + 1, cnt_non_zero + (digit != 0), is | (digit < limit));
  }
  return dp[pos][cnt_non_zero][is] = res;
}
long long solve() {
  memset(dp, -1, sizeof(dp));
  long long res = calc(0, 0, 0);
  for (int i = 1; i < 3; ++i) {
    if (dp[0][i][0] != -1) res += dp[0][i][0];
    if (dp[0][i][1] != -1) res += dp[0][i][1];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    s.clear();
    long long a, ans = 0;
    cin >> a;
    if (a == 1)
      s.push_back('0');
    else {
      --a;
      while (a) {
        s.push_back(a % 10 + '0');
        a /= 10LL;
      }
    }
    reverse(s.begin(), s.end());
    ans = solve();
    cin >> s;
    cout << solve() - ans << "\n";
  }
  return 0;
}