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;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
inline int Pow(int base, long long exp, int _mod) {
  if (!(base %= _mod)) return 0;
  int _ans = 1;
  for (; exp; exp >>= 1, base = (long long)base * base % _mod)
    exp& 1 ? _ans = (long long)_ans * base % _mod : 0;
  return _ans;
}
inline long long Pow(long long base, long long exp, long long _mod) {
  if (!(base %= _mod)) return 0;
  long long _ans = 1;
  for (; exp; exp >>= 1, base = base * base % _mod)
    exp& 1 ? _ans = _ans * base % _mod : 0;
  return _ans;
}
const int INF = 0x3f3f3f3f;
int c[20];
long long f(long long x) {
  if (x <= 0) return 0;
  long long xx = x;
  int len = 0;
  while (xx) c[++len] = xx % 10, xx /= 10;
  long long Ans[5], s;
  Ans[0] = Ans[1] = Ans[2] = Ans[3] = 0;
  s = 0;
  for (int i = len; i >= 1; --i) {
    Ans[3] = Ans[3] + Ans[2] * 9;
    Ans[2] = Ans[2] + Ans[1] * 9;
    Ans[1] = Ans[1] + Ans[0] * 9;
    if (c[i]) {
      if (s <= 3) ++Ans[s], Ans[s + 1] += c[i] - 1;
      ++s;
    }
  }
  return Ans[1] + Ans[2] + Ans[3] + (s <= 3);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long L, R;
    scanf("%lld%lld", &L, &R);
    printf("%lld\n", f(R) - f(L - 1));
  }
  return 0;
}