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;
long long l, r, lim[60];
long long f[60][60][2][2];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    (x *= 10) += ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long dfs(long long at, long long sum, bool is_lim, bool is_pre) {
  if (at == 0) return sum <= 3 && !is_pre;
  if (f[at][sum][is_lim][is_pre] != -1) return f[at][sum][is_lim][is_pre];
  long long mx = is_lim ? lim[at] : 9, res = 0;
  for (long long i = 0; i <= mx; i++) {
    if (i == 0 && is_pre)
      res += dfs(at - 1, sum, is_lim && i == mx, is_pre);
    else
      res += dfs(at - 1, sum + (i != 0), is_lim && i == mx, 0);
  }
  return f[at][sum][is_lim][is_pre] = res;
}
long long solve(long long x) {
  memset(f, -1, sizeof(f));
  memset(lim, 0, sizeof(lim));
  long long mx_pos = 0;
  while (x) lim[++mx_pos] = x % 10, x /= 10;
  return dfs(mx_pos, 0, 1, 1);
}
signed main() {
  long long T = read();
  while (T--) {
    l = read(), r = read();
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}