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 long long int N = 5005;
const long long int mod = 1e9 + 7;
const long long int level = 17;
long long int dx[] = {1, -1, 0, 0};
long long int dy[] = {0, 0, 1, -1};
vector<long long int> v;
long long int v_size;
long long int dp[19][2][4];
long long int func(long long int idx, long long int tight,
                   long long int non_zeros) {
  if (idx == v_size) return 1;
  if (dp[idx][tight][non_zeros] != -1) return dp[idx][tight][non_zeros];
  long long int ret = 0;
  if (tight == 1) {
    for (long long int i = 0; i <= v[idx]; i++) {
      long long int non_zero = 1;
      if (i == 0) non_zero = 0;
      if ((non_zero + non_zeros) <= 3) {
        if (i < v[idx])
          ret += func(idx + 1, 0, non_zero + non_zeros);
        else if (i == v[idx])
          ret += func(idx + 1, 1, non_zero + non_zeros);
      }
    }
  } else {
    for (long long int i = 0; i <= 9; i++) {
      long long int non_zero = 1;
      if (i == 0) non_zero = 0;
      if ((non_zero + non_zeros) <= 3) {
        ret += func(idx + 1, 0, non_zero + non_zeros);
      }
    }
  }
  return dp[idx][tight][non_zeros] = ret;
}
long long int solve(long long int x) {
  memset(dp, -1, sizeof(dp));
  if (x == 0) return 1;
  v.clear();
  while (x) {
    long long int rem = x % 10;
    x /= 10;
    v.push_back(rem);
  }
  reverse(v.begin(), v.end());
  v_size = v.size();
  long long int ans = func(0, 1, 0);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  long long int t;
  cin >> t;
  while (t--) {
    long long int l, r;
    cin >> l >> r;
    long long int ans = solve(r) - solve(l - 1);
    cout << ans << endl;
  }
  return 0;
}