#include <bits/stdc++.h>
using namespace std;
unsigned long long t, l, r, dp[21][21][21];
unsigned long long digit(unsigned long long i, unsigned long long n,
                         unsigned long long ok, unsigned long long nho,
                         unsigned long long a[]) {
  if (i == n + 1) {
    return 1;
  }
  unsigned long long res = dp[i][ok][nho];
  if (res != -1) return res;
  unsigned long long cnt = 0, block = 9;
  if (ok == 1) {
    block = a[i];
    for (int v = 0; v <= block; v++) {
      if (v < block) {
        unsigned long long nho1 = nho;
        if (v != 0) nho1++;
        if (nho1 > 3) break;
        cnt += digit(i + 1, n, 0, nho1, a);
      } else {
        long long nho1 = nho;
        if (v != 0) nho1++;
        if (nho1 > 3) break;
        cnt += digit(i + 1, n, 1, nho1, a);
      }
    }
  } else {
    for (int v = 0; v <= 9; v++) {
      unsigned long long nho1 = nho;
      if (v != 0) nho1++;
      if (nho1 > 3) break;
      cnt += digit(i + 1, n, 0, nho1, a);
    }
  }
  return dp[i][ok][nho] = cnt;
}
unsigned long long solve(unsigned long long x) {
  for (int i = 0; i <= 20; i++) {
    for (int j = 0; j <= 20; j++) {
      for (int k = 0; k <= 20; k++) {
        dp[i][j][k] = -1;
      }
    }
  }
  unsigned long long a[1000];
  unsigned long long tam = x, cnt = 0;
  while (tam != 0) {
    cnt++;
    tam /= 10;
  }
  unsigned long long c = cnt;
  while (x != 0) {
    a[cnt] = x % 10;
    x /= 10;
    cnt--;
  }
  return digit(1, c, 1, 0, a);
}
int main() {
  cin >> t;
  for (int i = 1; i <= t; i++) {
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

### 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;
unsigned long long t, l, r, dp[21][21][21];
unsigned long long digit(unsigned long long i, unsigned long long n,
                         unsigned long long ok, unsigned long long nho,
                         unsigned long long a[]) {
  if (i == n + 1) {
    return 1;
  }
  unsigned long long res = dp[i][ok][nho];
  if (res != -1) return res;
  unsigned long long cnt = 0, block = 9;
  if (ok == 1) {
    block = a[i];
    for (int v = 0; v <= block; v++) {
      if (v < block) {
        unsigned long long nho1 = nho;
        if (v != 0) nho1++;
        if (nho1 > 3) break;
        cnt += digit(i + 1, n, 0, nho1, a);
      } else {
        long long nho1 = nho;
        if (v != 0) nho1++;
        if (nho1 > 3) break;
        cnt += digit(i + 1, n, 1, nho1, a);
      }
    }
  } else {
    for (int v = 0; v <= 9; v++) {
      unsigned long long nho1 = nho;
      if (v != 0) nho1++;
      if (nho1 > 3) break;
      cnt += digit(i + 1, n, 0, nho1, a);
    }
  }
  return dp[i][ok][nho] = cnt;
}
unsigned long long solve(unsigned long long x) {
  for (int i = 0; i <= 20; i++) {
    for (int j = 0; j <= 20; j++) {
      for (int k = 0; k <= 20; k++) {
        dp[i][j][k] = -1;
      }
    }
  }
  unsigned long long a[1000];
  unsigned long long tam = x, cnt = 0;
  while (tam != 0) {
    cnt++;
    tam /= 10;
  }
  unsigned long long c = cnt;
  while (x != 0) {
    a[cnt] = x % 10;
    x /= 10;
    cnt--;
  }
  return digit(1, c, 1, 0, a);
}
int main() {
  cin >> t;
  for (int i = 1; i <= t; i++) {
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

```