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 <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
vector<long long> a;
long long pw[20];
void dfs(long long x, long long mul, long long k) {
  if (x == 18) {
    if (k) a.push_back(mul);
    return;
  }
  dfs(x + 1, mul, k);
  if (k == 3) return;
  for (long long i = 1; i <= 9; i++) dfs(x + 1, mul + pw[x] * i, k + 1);
}
void doit() {
  long long l, r;
  read(l, r);
  write(upper_bound(a.begin(), a.end(), r) -
        lower_bound(a.begin(), a.end(), l));
  puts("");
}
signed main() {
  pw[0] = 1;
  for (long long i = 1; i <= 18; i++) pw[i] = pw[i - 1] * 10ll;
  dfs(0, 0, 0);
  a.push_back(1e18);
  sort(a.begin(), a.end());
  long long t;
  read(t);
  while (t--) doit();
}