#include <bits/stdc++.h>
using namespace std;
int t, pop[1030];
long long l, r;
vector<long long> a;
void dfs(int pos, long long s, int cnt) {
  if (pos == 18) {
    a.push_back(s);
    return;
  }
  dfs(pos + 1, s * (long long)10, cnt);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      dfs(pos + 1, s * (long long)10 + i, cnt + 1);
    }
  }
  return;
}
int main() {
  dfs(0, 0, 0);
  a.push_back(1e18);
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &l, &r);
    printf("%d\n", upper_bound(a.begin(), a.end(), r) -
                       lower_bound(a.begin(), a.end(), l));
  }
}

### Prompt

Your task is to create 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, pop[1030];
long long l, r;
vector<long long> a;
void dfs(int pos, long long s, int cnt) {
  if (pos == 18) {
    a.push_back(s);
    return;
  }
  dfs(pos + 1, s * (long long)10, cnt);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      dfs(pos + 1, s * (long long)10 + i, cnt + 1);
    }
  }
  return;
}
int main() {
  dfs(0, 0, 0);
  a.push_back(1e18);
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &l, &r);
    printf("%d\n", upper_bound(a.begin(), a.end(), r) -
                       lower_bound(a.begin(), a.end(), l));
  }
}

```