#include <bits/stdc++.h>
using namespace std;
const long long N = 1e18;
vector<long long> q;
void spanning(long long p, long long t, long long v) {
  if (p == 18) return q.push_back(v), void();
  spanning(p + 1, t, v * 10);
  if (t >= 3) return;
  for (long long i = 1; i <= 9; i++) spanning(p + 1, t + 1, v * 10 + i);
}
long long calc(long long L, long long R) {
  return (long long)(upper_bound(q.begin(), q.end(), R) -
                     lower_bound(q.begin(), q.end(), L));
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
signed main() {
  spanning(0, 0, 0);
  q.push_back(N);
  long long t = read();
  while (t--) {
    long long L = read(), R = read();
    printf("%lld\n", calc(L, R));
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long N = 1e18;
vector<long long> q;
void spanning(long long p, long long t, long long v) {
  if (p == 18) return q.push_back(v), void();
  spanning(p + 1, t, v * 10);
  if (t >= 3) return;
  for (long long i = 1; i <= 9; i++) spanning(p + 1, t + 1, v * 10 + i);
}
long long calc(long long L, long long R) {
  return (long long)(upper_bound(q.begin(), q.end(), R) -
                     lower_bound(q.begin(), q.end(), L));
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
signed main() {
  spanning(0, 0, 0);
  q.push_back(N);
  long long t = read();
  while (t--) {
    long long L = read(), R = read();
    printf("%lld\n", calc(L, R));
  }
  return 0;
}

```