#include <bits/stdc++.h>
using namespace std;
long long int dp[66][20][20];
int bit[66];
long long int dfs(int pos, int num, int sum, bool lim) {
  if (sum > 3) return 0;
  if (pos == -1) {
    if (sum <= 3)
      return 1;
    else
      return 0;
  }
  if (!lim && dp[pos][num][sum] != -1) return dp[pos][num][sum];
  int up = lim ? bit[pos] : 9;
  long long int res = 0;
  for (int i = 0; i <= up; i++) {
    res += dfs(pos - 1, i, sum + (i > 0), lim && i == up);
  }
  if (!lim) dp[pos][num][sum] = res;
  return res;
}
long long int calc(long long int x) {
  int pos = 0;
  while (x) {
    bit[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 0, 1);
}
int main() {
  memset(dp, -1, sizeof(dp));
  int t;
  long long int x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &x, &y);
    long long int ans = calc(y) - calc(x - 1);
    printf("%lld\n", ans);
  }
  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;
long long int dp[66][20][20];
int bit[66];
long long int dfs(int pos, int num, int sum, bool lim) {
  if (sum > 3) return 0;
  if (pos == -1) {
    if (sum <= 3)
      return 1;
    else
      return 0;
  }
  if (!lim && dp[pos][num][sum] != -1) return dp[pos][num][sum];
  int up = lim ? bit[pos] : 9;
  long long int res = 0;
  for (int i = 0; i <= up; i++) {
    res += dfs(pos - 1, i, sum + (i > 0), lim && i == up);
  }
  if (!lim) dp[pos][num][sum] = res;
  return res;
}
long long int calc(long long int x) {
  int pos = 0;
  while (x) {
    bit[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 0, 1);
}
int main() {
  memset(dp, -1, sizeof(dp));
  int t;
  long long int x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &x, &y);
    long long int ans = calc(y) - calc(x - 1);
    printf("%lld\n", ans);
  }
  return 0;
}

```