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;
int lg2(const int &x) { return 31 - __builtin_clz(x); }
long long int lg2(const long long int &x) { return 63 - __builtin_clzll(x); }
char alpha[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                  'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                  's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
string s;
int t;
long long dp[20][50][2];
long long soln(int idx, int nz, int tyt) {
  if (idx == s.size()) {
    if (nz <= 3) return 1;
    return 0;
  }
  if (dp[idx][nz][tyt] != -1) return dp[idx][nz][tyt];
  long long res = 0;
  if (tyt) {
    for (int i = 0; i <= s[idx] - '0'; i++) {
      if (i == s[idx] - '0') {
        if (i == 0)
          res += soln(idx + 1, nz, 1);
        else
          res += soln(idx + 1, nz + 1, 1);
      } else {
        if (i == 0)
          res += soln(idx + 1, nz, 0);
        else
          res += soln(idx + 1, nz + 1, 0);
      }
    }
  } else {
    for (int i = 0; i <= 9; i++) {
      if (i == 0)
        res += soln(idx + 1, nz, 0);
      else
        res += soln(idx + 1, nz + 1, 0);
    }
  }
  return dp[idx][nz][tyt] = res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    memset(dp, -1, sizeof(dp));
    long long l, r;
    cin >> l >> r;
    s = to_string(r);
    long long cnt2 = soln(0, 0, 1);
    memset(dp, -1, sizeof(dp));
    s = to_string(l - 1);
    long long cnt1 = soln(0, 0, 1);
    cout << cnt2 - cnt1 << '\n';
  }
}