#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long dp[25][25];
int a[30];
int dfs(int pos, int sta, bool lead, bool limit) {
  if (pos == -1) return 1;
  if (!limit && !lead && dp[pos][sta] != -1) return dp[pos][sta];
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (lead && i == 0)
      ans += dfs(pos - 1, sta, lead, limit && i == a[pos]);
    else if (sta + (i != 0) <= 3)
      ans += dfs(pos - 1, sta + (i > 0 ? 1 : 0), 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, true, true);
}
int main() {
  memset(dp, -1, sizeof dp);
  int ww;
  scanf("%d", &ww);
  long long a, b;
  while (ww--) {
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", solve(b) - solve(a - 1));
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long dp[25][25];
int a[30];
int dfs(int pos, int sta, bool lead, bool limit) {
  if (pos == -1) return 1;
  if (!limit && !lead && dp[pos][sta] != -1) return dp[pos][sta];
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (lead && i == 0)
      ans += dfs(pos - 1, sta, lead, limit && i == a[pos]);
    else if (sta + (i != 0) <= 3)
      ans += dfs(pos - 1, sta + (i > 0 ? 1 : 0), 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, true, true);
}
int main() {
  memset(dp, -1, sizeof dp);
  int ww;
  scanf("%d", &ww);
  long long a, b;
  while (ww--) {
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", solve(b) - solve(a - 1));
  }
  return 0;
}

```