#include <bits/stdc++.h>
int Tim, num[20];
long long M[5][20];
long long run(int x, int cnt, int lim) {
  if (cnt > 3) return 0;
  if (!x) return 1;
  if ((~M[cnt][x]) && !lim) return M[cnt][x];
  long long ans = 0;
  for (int op = lim, i = lim ? num[x] : 9; ~i; i--, op = 0) {
    ans += run(x - 1, cnt + bool(i), op);
  }
  if (!lim) M[cnt][x] = ans;
  return ans;
}
long long solve(long long x) {
  int len = 0;
  while (x) {
    num[++len] = x % 10;
    x /= 10;
  }
  return run(len, 0, 1);
}
int main() {
  scanf("%d", &Tim);
  memset(M, -1, sizeof M);
  while (Tim-- > 0) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

### Prompt

Your task is to create 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>
int Tim, num[20];
long long M[5][20];
long long run(int x, int cnt, int lim) {
  if (cnt > 3) return 0;
  if (!x) return 1;
  if ((~M[cnt][x]) && !lim) return M[cnt][x];
  long long ans = 0;
  for (int op = lim, i = lim ? num[x] : 9; ~i; i--, op = 0) {
    ans += run(x - 1, cnt + bool(i), op);
  }
  if (!lim) M[cnt][x] = ans;
  return ans;
}
long long solve(long long x) {
  int len = 0;
  while (x) {
    num[++len] = x % 10;
    x /= 10;
  }
  return run(len, 0, 1);
}
int main() {
  scanf("%d", &Tim);
  memset(M, -1, sizeof M);
  while (Tim-- > 0) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

```