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;
int t, len = 19;
long long c[50][50], R[2], ans[2];
string sg[2];
void solve(int pos, int ct, string &r, long long &ans) {
  if (pos > len - 1 || ct < 0) return;
  solve(pos + 1, (r[pos] == '0' ? ct : ct - 1), r, ans);
  if (0) {
    if (ct)
      ans += r[pos] - '0' + 1;
    else
      ans++;
  } else
    for (int j = 0; j < r[pos] - '0'; j++) {
      for (int k = 0; k <= min((j ? ct - 1 : ct), len - 1 - pos); k++)
        ans += c[k][len - 1 - pos] * pow(9, k);
    }
  if (pos == len - 1 && ct && r[pos] != '0') ans++;
  if (pos == len - 1 && r[pos] == '0') ans++;
}
void combi() {
  c[0][0] = 1;
  for (int j = 1; j < 50; j++) {
    c[0][j] = 1;
    for (int i = 1; i <= j; i++) c[i][j] = c[i - 1][j - 1] + c[i][j - 1];
  }
}
int main() {
  combi();
  cin >> t;
  while (t--) {
    cin >> R[0] >> R[1];
    R[0]--;
    for (int i = 0; i < 2; i++) {
      string s(len - to_string(R[i]).length(), '0');
      ans[i] = 0;
      solve(0, 3, s.append(to_string(R[i])), ans[i]);
    }
    printf("%I64d\n", ans[1] - ans[0]);
  }
  return 0;
}