#include <bits/stdc++.h>
using namespace std;
int t, n;
long long L, R;
long long v[int(9e5)];
void solve(long long idx, int cnt, long long k) {
  if (idx == 18) {
    v[n++] = k;
    return;
  }
  solve(idx + 1, cnt, k * 10);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      solve(idx + 1, cnt + 1, k * 10 + i);
    }
  }
  return;
}
int bs(long long k) {
  int st = 0, en = n - 1;
  while (st < en) {
    int mid = (st + en + 1) >> 1;
    if (v[mid] <= k)
      st = mid;
    else
      en = mid - 1;
  }
  return st + 1;
}
int main() {
  scanf("%d", &t);
  solve(0, 0, 0);
  v[n++] = 1e18;
  sort(v, v + n);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    printf("%d\n", bs(R) - bs(L - 1));
  }
  return 0;
}

### Prompt

Generate 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>
using namespace std;
int t, n;
long long L, R;
long long v[int(9e5)];
void solve(long long idx, int cnt, long long k) {
  if (idx == 18) {
    v[n++] = k;
    return;
  }
  solve(idx + 1, cnt, k * 10);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      solve(idx + 1, cnt + 1, k * 10 + i);
    }
  }
  return;
}
int bs(long long k) {
  int st = 0, en = n - 1;
  while (st < en) {
    int mid = (st + en + 1) >> 1;
    if (v[mid] <= k)
      st = mid;
    else
      en = mid - 1;
  }
  return st + 1;
}
int main() {
  scanf("%d", &t);
  solve(0, 0, 0);
  v[n++] = 1e18;
  sort(v, v + n);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    printf("%d\n", bs(R) - bs(L - 1));
  }
  return 0;
}

```