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, lead;
long long c[50][50], ans;
string sg[2], s, t1 = "000000000000000000", t2 = "999999999999999999",
                 t3 = "1000000000000000000";
void solve(int pos, int ct, string &l, string &r) {
  if (pos > 17 || ct < 0) return;
  if (l[pos] == r[pos]) {
    if (l[pos] == '0')
      solve(pos + 1, ct, l, r);
    else
      solve(pos + 1, ct - 1, l, r);
    if (pos == 17 && ((l[pos] == '0') || (ct > 0))) ans++;
  } else {
    if (l[pos] == '0')
      solve(pos + 1, ct, l, t2);
    else
      solve(pos + 1, ct - 1, l, t2);
    solve(pos + 1, ct - 1, t1, r);
    if (pos < 17) {
      for (int k = 0; k <= min(ct - 1, 17 - pos); k++)
        ans += c[k][17 - pos] * 1LL * pow(9, k) * (r[pos] - l[pos] - 1);
    } else {
      if (ct == 0) {
        if (l[pos] == '0') ans++;
      } else
        ans += r[pos] - l[pos] + 1;
    }
  }
}
void combi() {
  c[0][1] = 1;
  c[1][1] = 1;
  for (int j = 2; 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 >> sg[0] >> sg[1];
    ans = 0;
    if (sg[0] == sg[1] && sg[1] == t3) {
      printf("1\n");
      continue;
    }
    if (sg[1] == t3) {
      ans++;
      sg[1] = t2;
    }
    for (int i = 0; i < 2; i++) {
      lead = 18 - sg[i].length();
      s = "";
      for (int j = 0; j < lead; j++) s = s + "0";
      s = s.append(sg[i]);
      sg[i] = s;
    }
    solve(0, 3, sg[0], sg[1]);
    printf("%I64d\n", ans);
  }
  return 0;
}