#include <bits/stdc++.h>
long long int dp(long long int now, long long int pos, int cnt) {
  if (pos == 0) return 1;
  if (cnt == 3) return 1;
  long long int ans = dp(now, pos / 10, cnt), p;
  for (p = 0; p < 10; p++)
    if (pos * p > now) break;
  p--;
  if (p > 0)
    ans += (p - 1) * dp(now - pos, pos / 10, cnt + 1) +
           dp(now - pos * p, pos / 10, cnt + 1);
  return ans;
}
int main() {
  int t;
  long long int l, r;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", dp(r, 1e18, 0) - dp(l - 1, 1e18, 0));
  }
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
long long int dp(long long int now, long long int pos, int cnt) {
  if (pos == 0) return 1;
  if (cnt == 3) return 1;
  long long int ans = dp(now, pos / 10, cnt), p;
  for (p = 0; p < 10; p++)
    if (pos * p > now) break;
  p--;
  if (p > 0)
    ans += (p - 1) * dp(now - pos, pos / 10, cnt + 1) +
           dp(now - pos * p, pos / 10, cnt + 1);
  return ans;
}
int main() {
  int t;
  long long int l, r;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", dp(r, 1e18, 0) - dp(l - 1, 1e18, 0));
  }
}

```