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>
const int maxn = 100001;
const double EPS = 1e-9;
const double PI = 3.141592653589793;
using namespace std;
const int SQRT = 350;
long long dp[20][2][5];
long long calc(long long val) {
  for (int i = 0; i < 20; ++i)
    for (int j = 0; j < 5; ++j) dp[i][0][j] = dp[i][1][j] = 0;
  string s = to_string(val);
  int n = s.size();
  for (int d = 0; d <= s[0] - '0'; ++d) {
    int ssss = 1;
    if (d == 0) ssss = 0;
    if (d == s[0] - '0')
      dp[1][1][ssss]++;
    else
      dp[1][0][ssss]++;
  }
  for (int len = 1; len < n; ++len) {
    for (int same = 0; same < 2; ++same)
      for (int cnt = 0; cnt <= 3; ++cnt) {
        for (int dig = 0; dig < 10; ++dig) {
          int good = 1;
          if (!dig) good = 0;
          if (!same && cnt + good <= 3) {
            dp[len + 1][0][cnt + good] += dp[len][0][cnt];
          }
          if (same && cnt + good <= 3 && dig <= s[len] - '0') {
            bool nsame = (dig == s[len] - '0');
            dp[len + 1][nsame][cnt + good] += dp[len][1][cnt];
          }
        }
      }
  }
  long long ans = 0;
  for (int i = 0; i < 4; ++i) {
    ans += dp[n][0][i];
    ans += dp[n][1][i];
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << calc(r) - calc(l - 1) << "\n";
  }
}