#include <bits/stdc++.h>
using namespace std;
long long a[30000000];
int m;
void dfs(int x, long long s, int k) {
  if (k > 3) return;
  if (x == 18) {
    a[m++] = s;
    return;
  }
  dfs(x + 1, s * 10, k);
  for (int i = 1; i <= 9; i++) {
    dfs(x + 1, s * 10 + i, k + 1);
  }
  return;
}
int main() {
  int t;
  long long x, y;
  m = 0;
  dfs(0, 0, 0);
  a[m++] = 1000000000000000000;
  sort(a, a + m);
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &x, &y);
    x = upper_bound(a, a + m, x - 1) - 1 - a;
    y = upper_bound(a, a + m, y) - 1 - a;
    printf("%lld\n", y - x);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long a[30000000];
int m;
void dfs(int x, long long s, int k) {
  if (k > 3) return;
  if (x == 18) {
    a[m++] = s;
    return;
  }
  dfs(x + 1, s * 10, k);
  for (int i = 1; i <= 9; i++) {
    dfs(x + 1, s * 10 + i, k + 1);
  }
  return;
}
int main() {
  int t;
  long long x, y;
  m = 0;
  dfs(0, 0, 0);
  a[m++] = 1000000000000000000;
  sort(a, a + m);
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &x, &y);
    x = upper_bound(a, a + m, x - 1) - 1 - a;
    y = upper_bound(a, a + m, y) - 1 - a;
    printf("%lld\n", y - x);
  }
  return 0;
}

```