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;
template <typename T>
inline void Cin(T &x) {
  char c;
  T sign = 1;
  x = 0;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sign;
}
const int N = 1e6 + 7;
const int M = 1e5 + 7;
const long double pi = 3.1415926535897;
long long f[20][2][20], l, q, ans;
string a;
stringstream cnv;
long long dp(int pos, bool prefix, int cnt) {
  if (pos >= a.size()) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (f[pos][prefix][cnt] > -1) return f[pos][prefix][cnt];
  int lim = prefix ? a[pos] - '0' : 9;
  long long res = 0;
  for (int i = 0; i <= lim; ++i) {
    if (i == 0)
      res += dp(pos + 1, prefix && (i == lim), cnt);
    else
      res += dp(pos + 1, prefix && (i == lim), cnt + 1);
  }
  return f[pos][prefix][cnt] = res;
}
void Enter() {
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    cin >> l >> a;
    --l;
    memset(&f, -1, sizeof f);
    ans = dp(0, 1, 0);
    cnv.clear();
    cnv << l;
    cnv >> a;
    memset(&f, -1, sizeof f);
    ans -= dp(0, 1, 0);
    cout << ans << '\n';
  }
}
void Solve() {}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Enter();
  Solve();
  return 0;
}