#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1024 * 4896, DIG = 18;
const long long INF = 1e18;
long long num[SIZE];
int n, cnt;
void init(int x, int y, long long z) {
  if (x == DIG) {
    num[++cnt] = z;
    return;
  }
  init(x + 1, y, z * 10);
  if (y < 3) {
    for (int i = 1; i <= 9; i++) {
      init(x + 1, y + 1, z * 10 + i);
    }
  }
}
inline int find(long long x) {
  int L = 1, R = cnt, M, ret;
  while (L <= R) {
    M = (L + R) / 2;
    if (num[M] <= x) {
      ret = M;
      L = M + 1;
    } else {
      R = M - 1;
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  num[++cnt] = INF;
  init(0, 0, 0);
  sort(num + 1, num + cnt + 1);
  for (int i = 1; i <= n; i++) {
    long long ii, jj;
    scanf("%lld%lld", &ii, &jj);
    printf("%d\n", find(jj) - find(ii - 1));
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int SIZE = 1024 * 4896, DIG = 18;
const long long INF = 1e18;
long long num[SIZE];
int n, cnt;
void init(int x, int y, long long z) {
  if (x == DIG) {
    num[++cnt] = z;
    return;
  }
  init(x + 1, y, z * 10);
  if (y < 3) {
    for (int i = 1; i <= 9; i++) {
      init(x + 1, y + 1, z * 10 + i);
    }
  }
}
inline int find(long long x) {
  int L = 1, R = cnt, M, ret;
  while (L <= R) {
    M = (L + R) / 2;
    if (num[M] <= x) {
      ret = M;
      L = M + 1;
    } else {
      R = M - 1;
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  num[++cnt] = INF;
  init(0, 0, 0);
  sort(num + 1, num + cnt + 1);
  for (int i = 1; i <= n; i++) {
    long long ii, jj;
    scanf("%lld%lld", &ii, &jj);
    printf("%d\n", find(jj) - find(ii - 1));
  }
  return 0;
}

```