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;
vector<int> num;
long long dp[405][250][2];
long long digitDP(int idx, long long nonZeroCnt, bool isSmall) {
  if (nonZeroCnt > 3) return 0;
  if (idx == (int)num.size()) return 1;
  if (dp[idx][nonZeroCnt][isSmall] != -1) return dp[idx][nonZeroCnt][isSmall];
  long long res = 0;
  int lmt = isSmall ? 9 : num[idx];
  for (int dgt = 0; dgt <= lmt; dgt++) {
    res +=
        digitDP(idx + 1, (dgt > 0) + nonZeroCnt, ((dgt < num[idx]) | isSmall));
  }
  return dp[idx][nonZeroCnt][isSmall] = res;
}
long long call(long long n) {
  num.clear();
  while (n > 0L) {
    num.push_back((int)(n % 10L));
    n /= 10L;
  }
  reverse(num.begin(), num.end());
  memset(dp, -1, sizeof dp);
  return digitDP(0, 0, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, l, r;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    cout << call(r) - call(l - 1) << '\n';
  }
  return 0;
}