#include <bits/stdc++.h>
using namespace std;
long long F[20][10];
long long num[20], len = 0;
long long dfs(long long ind, long long cnt, bool flag) {
  if (ind < 1) return 1;
  if (!flag && F[ind][cnt] != -1 && cnt <= 3) return F[ind][cnt];
  long long s = flag ? num[ind] : 9;
  long long ans = 0;
  ans += dfs(ind - 1, cnt, flag && s == 0);
  for (long long i = 1; i <= s; i++) {
    if (cnt >= 3) continue;
    ans += dfs(ind - 1, cnt + 1, i == s ? flag : 0);
  }
  if (!flag) F[ind][cnt] = ans;
  printf("");
  return ans;
}
long long solve(long long x) {
  memset(F, -1, sizeof(F));
  memset(num, 0, sizeof(num));
  len = 0;
  long long sum = x;
  while (sum) {
    num[++len] = sum % 10;
    sum /= 10;
  }
  long long ans = dfs(len, 0, 1);
  return ans;
}
int main() {
  long long T;
  scanf("%I64d", &T);
  while (T--) {
    long long L, R;
    scanf("%I64d%I64d", &L, &R);
    long long ans = solve(R) - solve(L - 1);
    printf("%I64d\n", ans);
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 F[20][10];
long long num[20], len = 0;
long long dfs(long long ind, long long cnt, bool flag) {
  if (ind < 1) return 1;
  if (!flag && F[ind][cnt] != -1 && cnt <= 3) return F[ind][cnt];
  long long s = flag ? num[ind] : 9;
  long long ans = 0;
  ans += dfs(ind - 1, cnt, flag && s == 0);
  for (long long i = 1; i <= s; i++) {
    if (cnt >= 3) continue;
    ans += dfs(ind - 1, cnt + 1, i == s ? flag : 0);
  }
  if (!flag) F[ind][cnt] = ans;
  printf("");
  return ans;
}
long long solve(long long x) {
  memset(F, -1, sizeof(F));
  memset(num, 0, sizeof(num));
  len = 0;
  long long sum = x;
  while (sum) {
    num[++len] = sum % 10;
    sum /= 10;
  }
  long long ans = dfs(len, 0, 1);
  return ans;
}
int main() {
  long long T;
  scanf("%I64d", &T);
  while (T--) {
    long long L, R;
    scanf("%I64d%I64d", &L, &R);
    long long ans = solve(R) - solve(L - 1);
    printf("%I64d\n", ans);
  }
}

```