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 long long MAX = 1e5 + 5;
const long long MOD = 1e9 + 7;
const long long inf = 1e8;
vector<long long> dig;
long long dp[25][5][2];
long long dfs(long long len, long long cnt, long long f) {
  if (cnt > 3) return 0;
  if (len == 0) return 1;
  if (dp[len][cnt][f] != -1) return dp[len][cnt][f];
  long long r = (f == 1) ? dig[len] : 9;
  long long ans = 0;
  for (int i = 0; i <= r; i++) {
    ans += dfs(len - 1, cnt + (i != 0), f & (i == dig[len]));
  }
  return dp[len][cnt][f] = ans;
}
long long solve(long long num) {
  memset(&dp, -1, sizeof dp);
  dig.resize(25, 0);
  long long len = 0;
  while (num != 0) {
    dig[++len] = num % 10;
    num /= 10;
  }
  long long ans = dfs(len, 0, 1);
  dig.clear();
  return ans;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    cout << solve(b) - solve(a - 1) << endl;
  }
  return 0;
}