#include <bits/stdc++.h>
using namespace std;
int t, n;
long long L, R, a, ans;
long long v[20];
long long dp[20][5][5];
long long solve(int idx, int cnt, bool p) {
  if (cnt > 3) return 0;
  if (idx == 0) return 1;
  long long &ret = dp[idx][cnt][p];
  if (ret != -1) return ret;
  int up = p ? 9 : v[idx];
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    ans += solve(idx - 1, cnt + (i > 0), (i != up) || p);
  }
  return ret = ans;
}
long long calc(long long X) {
  n = 0;
  while (X) {
    v[++n] = X % 10;
    X /= 10;
  }
  return solve(n, 0, 0);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    memset(dp, -1, sizeof(dp));
    ans = calc(R);
    memset(dp, -1, sizeof(dp));
    ans -= calc(L - 1);
    printf("%I64d\n", ans);
  }
  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, n;
long long L, R, a, ans;
long long v[20];
long long dp[20][5][5];
long long solve(int idx, int cnt, bool p) {
  if (cnt > 3) return 0;
  if (idx == 0) return 1;
  long long &ret = dp[idx][cnt][p];
  if (ret != -1) return ret;
  int up = p ? 9 : v[idx];
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    ans += solve(idx - 1, cnt + (i > 0), (i != up) || p);
  }
  return ret = ans;
}
long long calc(long long X) {
  n = 0;
  while (X) {
    v[++n] = X % 10;
    X /= 10;
  }
  return solve(n, 0, 0);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    memset(dp, -1, sizeof(dp));
    ans = calc(R);
    memset(dp, -1, sizeof(dp));
    ans -= calc(L - 1);
    printf("%I64d\n", ans);
  }
  return 0;
}

```