#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5;
vector<long long> vec;
void dfs(long long id, int cnt, int len) {
  vec.push_back(id);
  if (len == 18) return;
  dfs(id * 10, cnt, len + 1);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) dfs(id * 10 + i, cnt + 1, len + 1);
  }
}
void solve() {
  vec.clear();
  vec.push_back(1e18);
  for (int i = 1; i <= 9; i++) dfs(i, 1, 1);
  sort(vec.begin(), vec.end());
  int t;
  scanf("%d", &t);
  while (t--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long p = lower_bound(vec.begin(), vec.end(), l) - vec.begin();
    long long q = upper_bound(vec.begin(), vec.end(), r) - vec.begin();
    printf("%lld\n", q - p);
  }
}
int main() { solve(); }

### 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;
const int maxn = 3e5;
vector<long long> vec;
void dfs(long long id, int cnt, int len) {
  vec.push_back(id);
  if (len == 18) return;
  dfs(id * 10, cnt, len + 1);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) dfs(id * 10 + i, cnt + 1, len + 1);
  }
}
void solve() {
  vec.clear();
  vec.push_back(1e18);
  for (int i = 1; i <= 9; i++) dfs(i, 1, 1);
  sort(vec.begin(), vec.end());
  int t;
  scanf("%d", &t);
  while (t--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long p = lower_bound(vec.begin(), vec.end(), l) - vec.begin();
    long long q = upper_bound(vec.begin(), vec.end(), r) - vec.begin();
    printf("%lld\n", q - p);
  }
}
int main() { solve(); }

```