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;
const int MAXN = 200200;
const int MAXE = 400400;
const int INF = 0x3f3f3f3f;
template <typename T>
inline void read(T &x) {
  int ch = getchar();
  bool fg = false;
  for (x = 0; !isdigit(ch); ch = getchar()) {
    if (ch == '-') {
      fg = true;
    }
  }
  for (; isdigit(ch); ch = getchar()) {
    x = x * 10 + ch - '0';
  }
  if (fg) {
    x = -x;
  }
}
long long T, L, R;
long long f[MAXN * 100], cnt;
void dfs(int fg, long long now) {
  if (now > 1000000000000000000LL || now < 0) return;
  f[++cnt] = now;
  dfs(fg, now * 10);
  if (fg < 3)
    for (int i = 1; i <= 9; i++) {
      dfs(fg + 1, now * 10 + i);
    }
}
signed main() {
  read(T);
  for (int i = 1; i <= 9; i++) dfs(1, i);
  sort(f + 1, f + cnt + 1);
  cnt = (long long)(unique(f + 1, f + cnt + 1) - f - 1);
  while (T--) {
    read(L), read(R);
    printf("%I64d\n", (long long)(upper_bound(f + 1, f + cnt + 1, R) -
                                  lower_bound(f + 1, f + cnt + 1, L)));
  }
  return 0;
}