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;
using LL = long long;
int a[25];
LL dp[25][5], l, r;
LL dfs(int pos, int cnt, bool zero, bool limit) {
  if (pos == -1) return 1;
  if (!limit && !zero && dp[pos][cnt] != -1) return dp[pos][cnt];
  int up = limit ? a[pos] : 9;
  if (cnt == 3) up = 0;
  LL ans = 0;
  for (int i = 0; i <= up; ++i)
    ans += dfs(pos - 1, cnt + (i == 0 ? 0 : 1), zero && i == 0,
               limit && i == a[pos]);
  if (!limit && !zero) dp[pos][cnt] = ans;
  return ans;
}
LL solve(LL x) {
  int pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, true, true);
}
int main() {
  int T;
  LL l, r;
  scanf("%d", &T);
  while (T--) {
    memset(dp, -1, sizeof dp);
    scanf("%I64d%I64d", &l, &r);
    printf("%I64d\n", solve(r) - solve(l - 1));
  }
  return 0;
}