#include <bits/stdc++.h>
using namespace std;
long long dp[25][2][4];
int digit[25];
long long fun(int len, int sum, int limit) {
  if (len < 0) return 1;
  if (dp[len][limit][sum]) return dp[len][limit][sum];
  if (sum >= 3) return dp[len][limit][sum] = 1;
  long long cnt = 0;
  int p = limit ? digit[len] : 9;
  for (int i = 0; i <= p; i++) {
    cnt += fun(len - 1, sum + (i == 0 ? 0 : 1), (i == p) && limit);
  }
  return dp[len][limit][sum] = cnt;
}
long long solve(long long n) {
  int k = 0;
  memset(dp, 0, sizeof dp);
  while (n) {
    digit[k++] = n % 10;
    n /= 10;
  }
  return fun(k - 1, 0, 1);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 dp[25][2][4];
int digit[25];
long long fun(int len, int sum, int limit) {
  if (len < 0) return 1;
  if (dp[len][limit][sum]) return dp[len][limit][sum];
  if (sum >= 3) return dp[len][limit][sum] = 1;
  long long cnt = 0;
  int p = limit ? digit[len] : 9;
  for (int i = 0; i <= p; i++) {
    cnt += fun(len - 1, sum + (i == 0 ? 0 : 1), (i == p) && limit);
  }
  return dp[len][limit][sum] = cnt;
}
long long solve(long long n) {
  int k = 0;
  memset(dp, 0, sizeof dp);
  while (n) {
    digit[k++] = n % 10;
    n /= 10;
  }
  return fun(k - 1, 0, 1);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

```