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 l, r;
string s;
long long dp[20][5][2];
int solve(long long pos, long long cnt, long long tight) {
  if (pos == s.length()) return 1;
  if (dp[pos][cnt][tight] != -1) {
    return dp[pos][cnt][tight];
  }
  long long end = tight ? s[pos] - '0' : 9;
  long long ans = 0;
  for (int i = 0; i <= end; i++) {
    int cntupd = i > 0 ? cnt + 1 : cnt;
    if (cntupd <= 3) {
      ans += solve(pos + 1, cntupd, tight & (i == end));
    }
  }
  return dp[pos][cnt][tight] = ans;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    s = to_string(r);
    memset(dp, -1, sizeof(dp));
    long long ans = solve(0, 0, 1);
    l--;
    memset(dp, -1, sizeof(dp));
    s = to_string(l);
    ans -= solve(0, 0, 1);
    cout << ans << endl;
  }
}