#include <bits/stdc++.h>
using namespace std;
int t;
long long l, r, w;
long long tab[2][2];
int dg(long long a) {
  int ans = 0;
  while (a) {
    if (a % 10) ans++;
    a /= 10;
  }
  return ans;
}
long long f(int k, int n) {
  if (k > 3) return 0;
  if (k == 3) return 1;
  if (k == 2) return (9 * n) + 1;
  if (k == 1) return (81 * n * (n - 1) / 2) + (9 * n) + 1;
}
void nr(long long a, bool b) {
  int j = 0;
  long long d = 10, a2 = a, lim = 1;
  while (a2) {
    a2 /= 10;
    j++;
  }
  tab[b][0] = j;
  if (j == 19) return;
  for (int i = 0; i < j; i++) lim *= 10;
  j = 0;
  while (a > d / 10 && a < lim) {
    if (10 - ((a % d) * 10 / d) == 10)
      tab[b][1] += f(dg(a), j) + 9 * f(dg(a) + 1, j);
    else
      tab[b][1] += (10 - ((a % d) * 10 / d)) * f(dg(a), j);
    a += (10 - ((a % d) * 10 / d)) * d / 10;
    d *= 10;
    j++;
  }
  if (a < 10) tab[b][1] = 10 - a;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> l >> r;
    w = 0;
    tab[0][1] = 0;
    tab[1][1] = 0;
    nr(l, 0);
    nr(r, 1);
    for (int j = tab[0][0] + 1; j <= min(tab[1][0], 18ll); j++) {
      w += 9 * f(1, j - 1);
    }
    cout << w - tab[1][1] + tab[0][1] + (dg(r) < 4) << "\n";
  }
  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;
int t;
long long l, r, w;
long long tab[2][2];
int dg(long long a) {
  int ans = 0;
  while (a) {
    if (a % 10) ans++;
    a /= 10;
  }
  return ans;
}
long long f(int k, int n) {
  if (k > 3) return 0;
  if (k == 3) return 1;
  if (k == 2) return (9 * n) + 1;
  if (k == 1) return (81 * n * (n - 1) / 2) + (9 * n) + 1;
}
void nr(long long a, bool b) {
  int j = 0;
  long long d = 10, a2 = a, lim = 1;
  while (a2) {
    a2 /= 10;
    j++;
  }
  tab[b][0] = j;
  if (j == 19) return;
  for (int i = 0; i < j; i++) lim *= 10;
  j = 0;
  while (a > d / 10 && a < lim) {
    if (10 - ((a % d) * 10 / d) == 10)
      tab[b][1] += f(dg(a), j) + 9 * f(dg(a) + 1, j);
    else
      tab[b][1] += (10 - ((a % d) * 10 / d)) * f(dg(a), j);
    a += (10 - ((a % d) * 10 / d)) * d / 10;
    d *= 10;
    j++;
  }
  if (a < 10) tab[b][1] = 10 - a;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> l >> r;
    w = 0;
    tab[0][1] = 0;
    tab[1][1] = 0;
    nr(l, 0);
    nr(r, 1);
    for (int j = tab[0][0] + 1; j <= min(tab[1][0], 18ll); j++) {
      w += 9 * f(1, j - 1);
    }
    cout << w - tab[1][1] + tab[0][1] + (dg(r) < 4) << "\n";
  }
  return 0;
}

```