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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, long long p, long long MOD) {
  if (p == 0) return 1 % MOD;
  if (p == 1) return x % MOD;
  long long res = power(x, p / 2, MOD);
  res = (long long)res * res % MOD;
  if (p & 1) res = (long long)res * x % MOD;
  return res;
}
vector<long long> repr(long long x) {
  vector<long long> v;
  while (x) {
    v.push_back(x % 10);
    x /= 10;
  }
  reverse((v).begin(), (v).end());
  return v;
}
const long long N = 30;
long long dp[N][4][2];
vector<long long> r;
long long n;
long long go(long long i, long long cnt, bool f) {
  if (i == n) return 1;
  long long& res = dp[i][cnt][f];
  if (res != -1) return res;
  res = 0;
  long long til = f ? 9 : r[i];
  for (long long j = 0; j <= til; j++) {
    if (!j || cnt < 3) {
      res += go(i + 1, cnt + (j > 0), f || j < r[i]);
    }
  }
  return res;
}
long long solve(long long x) {
  if (!x) return 1;
  r = repr(x);
  n = r.size();
  memset((dp), (-1), sizeof(dp));
  return go(0, 0, false);
}
int32_t main() {
  long long T;
  cin >> T;
  while (T--) {
    long long a, b;
    cin >> a >> b;
    cout << solve(b) - solve(a - 1) << endl;
  }
}