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;
const int MAXN = 25;
int a[MAXN][4], b[MAXN];
long long solve(long long x) {
  int n = 0;
  long long ret = 0;
  if (x == 0) return 1;
  while (x) b[n++] = x % 10, x /= 10;
  for (int i = n - 1, nonz = 0; i >= 0 && nonz <= 3; --i)
    if (i == 0) {
      if (nonz < 3)
        ret += b[i] + 1;
      else
        ++ret;
    } else {
      if (b[i] > 0) ret += a[i][3 - nonz];
      if (nonz < 3 && b[i] > 1) ret += (b[i] - 1) * a[i][2 - nonz];
      nonz += (b[i] > 0);
    }
  return ret;
}
int main() {
  int tn;
  long long l, r;
  for (int i = 0; i < MAXN; ++i) {
    a[i][0] = 1;
    a[i][1] = i * 9;
    a[i][2] = i * (i - 1) / 2 * 9 * 9;
    a[i][3] = i * (i - 1) * (i - 2) / 6 * 9 * 9 * 9;
    for (int j = 1; j <= 3; ++j) a[i][j] += a[i][j - 1];
  }
  scanf("%d", &tn);
  for (int i = 0; i < tn; ++i)
    scanf("%lld%lld", &l, &r), printf("%lld\n", solve(r) - solve(l - 1));
  return 0;
}