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 D = 20;
int n, a[D];
long long dp[D][4];
long long calc(long long x) {
  if (!x) return 0;
  n = 0;
  while (x) {
    a[++n] = x % 10;
    x /= 10;
  }
  int sum = 0, z;
  long long sol = 0;
  for (int i = n; i; i--) {
    for (int j = 0; j < a[i]; j++) {
      if (j)
        z = sum + 1;
      else
        z = sum;
      if (z <= 3) sol += dp[i - 1][3 - z];
    }
    if (a[i]) sum++;
  }
  return sol;
}
int main() {
  int q;
  scanf("%d", &q);
  dp[0][0] = dp[0][1] = dp[0][2] = dp[0][3] = 1;
  for (int i = 1; i < D; i++)
    for (int j = 0; j <= 3; j++) {
      if (j) dp[i][j] = 9 * dp[i - 1][j - 1];
      dp[i][j] += dp[i - 1][j];
    }
  while (q--) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    printf("%lld\n", calc(y + 1) - calc(x));
  }
  return 0;
}