#include <bits/stdc++.h>
using namespace std;
long long dp[60][5];
long long a[30];
long long dfs(int pos, int pre, bool limit) {
  if (pre > 3) return 0;
  if (pos == -1 && pre <= 3)
    return 1;
  else if (pos == -1)
    return 0;
  if (!limit && dp[pos][pre] != -1) return dp[pos][pre];
  int up = limit ? a[pos] : 9;
  int sum = 0;
  for (int i = 0; i <= up; i++) {
    sum += dfs(pos - 1, pre + (i != 0), limit && i == a[pos]);
  }
  if (!limit) dp[pos][pre] = sum;
  return sum;
}
long long solve(long long k) {
  int pos = 0;
  while (k) {
    a[pos++] = k % 10;
    k /= 10;
  }
  long long ans = dfs(pos - 1, 0, 1);
  return ans;
}
int main() {
  int t;
  long long n, m;
  scanf("%d", &t);
  memset(dp, -1, sizeof(dp));
  while (t--) {
    memset(a, 0, sizeof(a));
    scanf("%lld%lld", &n, &m);
    printf(t == 0 ? "%lld" : "%lld\n", solve(m) - solve(n - 1));
  }
  return 0;
}

### Prompt

Generate a Cpp solution 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;
long long dp[60][5];
long long a[30];
long long dfs(int pos, int pre, bool limit) {
  if (pre > 3) return 0;
  if (pos == -1 && pre <= 3)
    return 1;
  else if (pos == -1)
    return 0;
  if (!limit && dp[pos][pre] != -1) return dp[pos][pre];
  int up = limit ? a[pos] : 9;
  int sum = 0;
  for (int i = 0; i <= up; i++) {
    sum += dfs(pos - 1, pre + (i != 0), limit && i == a[pos]);
  }
  if (!limit) dp[pos][pre] = sum;
  return sum;
}
long long solve(long long k) {
  int pos = 0;
  while (k) {
    a[pos++] = k % 10;
    k /= 10;
  }
  long long ans = dfs(pos - 1, 0, 1);
  return ans;
}
int main() {
  int t;
  long long n, m;
  scanf("%d", &t);
  memset(dp, -1, sizeof(dp));
  while (t--) {
    memset(a, 0, sizeof(a));
    scanf("%lld%lld", &n, &m);
    printf(t == 0 ? "%lld" : "%lld\n", solve(m) - solve(n - 1));
  }
  return 0;
}

```