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;
void shoot() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long power(long long n, long long pwr) {
  long long ans = 1LL;
  while (pwr) {
    if (pwr & 1) ans = (ans * n);
    n = (n * n);
    pwr >>= 1;
  }
  return ans;
}
long long nCr(long long i, long long j) {
  if (j == 0) return 1;
  if (j == 1) return i;
  if (j == 2) return (i * (i - 1)) / 2;
  return (i * (i - 1) * (i - 2)) / 6;
}
long long go(int n, int k) {
  long long i, sum = 0;
  for (i = 0; i <= k; i++) {
    sum += nCr(n, i) * power(9, i);
  }
  return sum;
}
int main() {
  shoot();
  long long t, i, j;
  cin >> t;
  while (t--) {
    long long i, l, r, n, cur, ans = 0;
    cin >> l >> r;
    string s = to_string(r + 1);
    ans = 0;
    cur = 3;
    n = s.size();
    for (i = 0; i < n; i++) {
      if (s[i] == '0') continue;
      ans += go(n - i - 1, cur);
      --cur;
      if (cur == -1) break;
      ans += go(n - i - 1, cur) * (s[i] - '1');
    }
    s = to_string(l);
    cur = 3;
    n = s.size();
    for (i = 0; i < n; i++) {
      if (s[i] == '0') continue;
      ans -= go(n - i - 1, cur);
      --cur;
      if (cur == -1) break;
      ans -= go(n - i - 1, cur) * (s[i] - '1');
    }
    cout << ans << endl;
  }
  return 0;
}