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>
bool SUBMIT = 1;
using namespace std;
const double PI = acos(-1);
long long n, k[20], sn[20][2][4];
long long dp(long long l, int up, long long check) {
  if (l < 0) return 1;
  long long c = sn[l][up][check];
  if (c) return c;
  long long ans = 0;
  if (check >= 3) return sn[l][up][check] = 1;
  long long m;
  if (up)
    m = k[l];
  else
    m = 9;
  for (int i = 0; i <= m; i++) {
    ans += dp(l - 1, up && (i == m), check + (i == 0 ? 0 : 1));
  }
  return sn[l][up][check] = ans;
}
long long fun(long long m) {
  long long l = 0;
  memset(sn, 0, sizeof(sn));
  while (m > 0) {
    k[l++] = m % 10;
    m /= 10;
  }
  return dp(l - 1, 1, 0);
}
int main() {
  if (!SUBMIT)
    freopen("i.txt", "r", stdin);
  else
    ios::sync_with_stdio(false);
  cin >> n;
  while (n--) {
    long long a, b;
    cin >> a >> b;
    long long ans = fun(b) - fun(a - 1);
    cout << ans << '\n';
  }
  return 0;
}