#include <bits/stdc++.h>
using namespace std;
long long dp[25][4], a[25];
long long dfs(int pos, int sta, int lim) {
  if (pos == -1) return 1;
  if (!lim && ~dp[pos][sta]) return dp[pos][sta];
  int sup = lim ? a[pos] : 9;
  long long tmp = 0;
  for (int i = 0; i <= sup; i++) {
    if (i && sta == 3) continue;
    tmp += dfs(pos - 1, sta + (i ? 1 : 0), lim & (i == a[pos]));
  }
  if (!lim) dp[pos][sta] = tmp;
  return tmp;
}
long long solve(long long x) {
  int p = -1;
  for (; x; x /= 10) a[++p] = x % 10;
  return dfs(p, 0, 1);
}
int main() {
  int T;
  cin >> T;
  memset(dp, -1, sizeof dp);
  while (T--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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[25][4], a[25];
long long dfs(int pos, int sta, int lim) {
  if (pos == -1) return 1;
  if (!lim && ~dp[pos][sta]) return dp[pos][sta];
  int sup = lim ? a[pos] : 9;
  long long tmp = 0;
  for (int i = 0; i <= sup; i++) {
    if (i && sta == 3) continue;
    tmp += dfs(pos - 1, sta + (i ? 1 : 0), lim & (i == a[pos]));
  }
  if (!lim) dp[pos][sta] = tmp;
  return tmp;
}
long long solve(long long x) {
  int p = -1;
  for (; x; x /= 10) a[++p] = x % 10;
  return dfs(p, 0, 1);
}
int main() {
  int T;
  cin >> T;
  memset(dp, -1, sizeof dp);
  while (T--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

```