#include <bits/stdc++.h>
using namespace std;
const int N = 20;
long long dp[N][5];
long long a[N];
long long dfs(int pos, int pre, bool limit) {
  if (pre > 3) return 0;
  if (pos == -1) {
    if (pre <= 3)
      return 1;
    else
      return 0;
  }
  if (!limit && dp[pos][pre] != -1) return dp[pos][pre];
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (i == 0)
      ans += dfs(pos - 1, pre, limit && i == a[pos]);
    else
      ans += dfs(pos - 1, pre + 1, limit && i == a[pos]);
  }
  if (!limit) dp[pos][pre] = ans;
  return ans;
}
long long f(long long x) {
  int i = 0;
  while (x) {
    a[i++] = x % 10;
    x /= 10;
  }
  return dfs(i - 1, 0, 1);
}
int main() {
  int tt;
  scanf("%d", &tt);
  long long l, r;
  memset(dp, -1, sizeof(dp));
  while (tt--) {
    scanf("%I64d%I64d", &l, &r);
    long long ans1 = f(r);
    long long ans2 = f(l - 1);
    printf("%I64d\n", ans1 - ans2);
  }
}

### 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;
const int N = 20;
long long dp[N][5];
long long a[N];
long long dfs(int pos, int pre, bool limit) {
  if (pre > 3) return 0;
  if (pos == -1) {
    if (pre <= 3)
      return 1;
    else
      return 0;
  }
  if (!limit && dp[pos][pre] != -1) return dp[pos][pre];
  int up = limit ? a[pos] : 9;
  long long ans = 0;
  for (int i = 0; i <= up; i++) {
    if (i == 0)
      ans += dfs(pos - 1, pre, limit && i == a[pos]);
    else
      ans += dfs(pos - 1, pre + 1, limit && i == a[pos]);
  }
  if (!limit) dp[pos][pre] = ans;
  return ans;
}
long long f(long long x) {
  int i = 0;
  while (x) {
    a[i++] = x % 10;
    x /= 10;
  }
  return dfs(i - 1, 0, 1);
}
int main() {
  int tt;
  scanf("%d", &tt);
  long long l, r;
  memset(dp, -1, sizeof(dp));
  while (tt--) {
    scanf("%I64d%I64d", &l, &r);
    long long ans1 = f(r);
    long long ans2 = f(l - 1);
    printf("%I64d\n", ans1 - ans2);
  }
}

```