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 a[20];
long long dp[20][5];
long long dfs(int pos, int num, bool lead, bool limit) {
  if (pos == -1 && num <= 3) return 1;
  if (!limit && !lead && !limit && dp[pos][num] != -1) return dp[pos][num];
  if (num > 3) return 0;
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (i != 0)
      ans += dfs(pos - 1, num + 1, 0, limit && i == a[pos]);
    else
      ans += dfs(pos - 1, num, 1, limit && i == a[pos]);
  }
  if (!limit && !lead) dp[pos][num] = ans;
  return ans;
}
long long solve(long long num) {
  int pos = 0;
  while (num) {
    a[pos++] = num % 10;
    num /= 10;
  }
  return dfs(pos - 1, 0, true, true);
}
int main() {
  memset(dp, -1, sizeof(dp));
  int T;
  scanf("%d", &T);
  while (T--) {
    long long L, R;
    scanf("%lld %lld", &L, &R);
    printf("%lld\n", solve(R) - solve(L - 1));
  }
  return 0;
}