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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long max3(long long x, long long y, long long z) {
  return max(max(x, y), z);
}
long long min3(long long x, long long y, long long z) {
  return min(min(x, y), z);
}
const long long N = 1e5 + 10, M = 2e5 + 10, M2 = 1e6 + 10, mod = 1e9 + 7,
                inf = 1e17 + 10;
const long long INF = 1e9 + 7;
void add(long long& a, long long b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
long long X[] = {0, 1, 0, -1};
long long Y[] = {-1, 0, 1, 0};
long long power(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = ((x % mod) * (x % mod)) % mod;
    n = n / 2;
  }
  return result;
}
vector<long long> classy;
void ye_te_brute(long long pos, long long cnt, long long cur) {
  if (pos == 18) {
    classy.push_back(cur);
    return;
  }
  ye_te_brute(pos + 1, cnt, cur * 10);
  if (cnt < 3) {
    for (long long i = 1; i <= 9; i++) {
      ye_te_brute(pos + 1, cnt + 1, cur * 10 + i);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  ye_te_brute(0, 0, 0);
  classy.push_back(1000000000000000000);
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long res1 =
        upper_bound(classy.begin(), classy.end(), r) - classy.begin();
    long long res2 =
        lower_bound(classy.begin(), classy.end(), l) - classy.begin();
    cout << res1 - res2 << "\n";
  }
  return 0;
}