#include <bits/stdc++.h>
using namespace std;
long long d[99][9];
void init() {
  d[0][0] = 1;
  for (int i = 1; i <= 18; i++) {
    d[i][0] = 1;
    d[i][1] = 9 + d[i - 1][1];
    d[i][2] = d[i - 1][2] + d[i - 1][1] * 9;
    d[i][3] = d[i - 1][3] + d[i - 1][2] * 9;
  }
}
int a[99];
long long calc(long long R) {
  if (R == 0) return 0;
  a[0] = 0;
  long long tem = R;
  while (tem) a[++a[0]] = tem % 10, tem /= 10;
  long long ret = 0;
  int num = 0;
  for (int i = a[0]; i >= 1; i--) {
    if (!a[i]) continue;
    if (num == 0)
      ret += (a[i] - 1) * (d[i - 1][2] + d[i - 1][1] + 1) + d[i - 1][3] +
             d[i - 1][2] + d[i - 1][1];
    if (num == 1)
      ret += (a[i] - 1) * (d[i - 1][1] + 1) + d[i - 1][2] + d[i - 1][1] + 1;
    if (num == 2) ret += a[i] + d[i - 1][1] + 1;
    ++num;
  }
  if (num < 3) ret++;
  return ret;
}
int main() {
  init();
  int T;
  scanf("%d", &T);
  while (T--) {
    long long L, R;
    scanf("%I64d%I64d", &L, &R);
    printf("%I64d\n", calc(R) - calc(L - 1));
  }
  return 0;
}

### Prompt

Your challenge is to write 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 d[99][9];
void init() {
  d[0][0] = 1;
  for (int i = 1; i <= 18; i++) {
    d[i][0] = 1;
    d[i][1] = 9 + d[i - 1][1];
    d[i][2] = d[i - 1][2] + d[i - 1][1] * 9;
    d[i][3] = d[i - 1][3] + d[i - 1][2] * 9;
  }
}
int a[99];
long long calc(long long R) {
  if (R == 0) return 0;
  a[0] = 0;
  long long tem = R;
  while (tem) a[++a[0]] = tem % 10, tem /= 10;
  long long ret = 0;
  int num = 0;
  for (int i = a[0]; i >= 1; i--) {
    if (!a[i]) continue;
    if (num == 0)
      ret += (a[i] - 1) * (d[i - 1][2] + d[i - 1][1] + 1) + d[i - 1][3] +
             d[i - 1][2] + d[i - 1][1];
    if (num == 1)
      ret += (a[i] - 1) * (d[i - 1][1] + 1) + d[i - 1][2] + d[i - 1][1] + 1;
    if (num == 2) ret += a[i] + d[i - 1][1] + 1;
    ++num;
  }
  if (num < 3) ret++;
  return ret;
}
int main() {
  init();
  int T;
  scanf("%d", &T);
  while (T--) {
    long long L, R;
    scanf("%I64d%I64d", &L, &R);
    printf("%I64d\n", calc(R) - calc(L - 1));
  }
  return 0;
}

```