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;
const int MAX = (int)1e5 + 55;
const int INF = (int)1e9 + 77;
const int MOD = (int)1e9 + 7;
const double PI = 2 * acos(0.0);
const double EPS = 1e-7;
vector<long long> v;
void build(int idx, int cnt, long long num) {
  if (idx == 19 || cnt > 3) return;
  v.push_back(num);
  for (int i = 0; i <= 9; ++i) build(idx + 1, cnt + (i != 0), num * 10 + i);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i <= 9; ++i) build(1, 1, i);
  v.push_back(0);
  v.push_back((long long)1e18);
  sort(v.begin(), v.end());
  int tc;
  cin >> tc;
  while (tc--) {
    long long l, r;
    cin >> l >> r;
    cout << upper_bound(v.begin(), v.end(), r) -
                lower_bound(v.begin(), v.end(), l)
         << '\n';
  }
  return 0;
}