#include <bits/stdc++.h>
using namespace std;
int a[20];
long long dp[20][20];
long long dfs(int pos, int sta, bool lead, int limit) {
  if (pos == -1) return sta <= 3;
  if (!limit && !lead && dp[pos][sta] != -1) return dp[pos][sta];
  int len = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= len; i++) {
    ans += dfs(pos - 1, sta + (i == 0 ? 0 : 1), lead && i == 0,
               limit && i == a[pos]);
  }
  if (!limit && !lead) dp[pos][sta] = ans;
  return ans;
}
long long solve(long long x) {
  int pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 1, 1);
}
int main() {
  long long a, b;
  int t;
  cin >> t;
  memset(dp, -1, sizeof(dp));
  while (t--) {
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", solve(b) - solve(a - 1));
  }
}

### Prompt

In Cpp, your task is to solve 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 a[20];
long long dp[20][20];
long long dfs(int pos, int sta, bool lead, int limit) {
  if (pos == -1) return sta <= 3;
  if (!limit && !lead && dp[pos][sta] != -1) return dp[pos][sta];
  int len = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= len; i++) {
    ans += dfs(pos - 1, sta + (i == 0 ? 0 : 1), lead && i == 0,
               limit && i == a[pos]);
  }
  if (!limit && !lead) dp[pos][sta] = ans;
  return ans;
}
long long solve(long long x) {
  int pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 1, 1);
}
int main() {
  long long a, b;
  int t;
  cin >> t;
  memset(dp, -1, sizeof(dp));
  while (t--) {
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", solve(b) - solve(a - 1));
  }
}

```