#include <bits/stdc++.h>
using namespace std;
long long int dp[19][4][2];
string s;
long long int solve(long long int i, long long int cnt, long long int f) {
  if (cnt > 3) {
    return 0;
  }
  if (i == s.size()) {
    return 1;
  }
  long long int ans = 0;
  if (dp[i][cnt][f] != -1) {
    return dp[i][cnt][f];
  }
  for (long long int j = 0; j < 10; j++) {
    if (j < s[i] - '0') {
      ans += solve(i + 1, cnt + !!j, 1);
    } else if (f || s[i] - '0' == j) {
      ans += solve(i + 1, cnt + !!j, f);
    }
  }
  dp[i][cnt][f] = ans;
  return dp[i][cnt][f];
}
int main() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    long long int l, r;
    cin >> l >> r;
    l--;
    s = to_string(r);
    long long int ans = 0;
    memset(dp, -1, sizeof(dp));
    ans += solve(0, 0, 0);
    memset(dp, -1, sizeof(dp));
    s = to_string(l);
    ans -= solve(0, 0, 0);
    cout << ans << '\n';
  }
}

### 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;
long long int dp[19][4][2];
string s;
long long int solve(long long int i, long long int cnt, long long int f) {
  if (cnt > 3) {
    return 0;
  }
  if (i == s.size()) {
    return 1;
  }
  long long int ans = 0;
  if (dp[i][cnt][f] != -1) {
    return dp[i][cnt][f];
  }
  for (long long int j = 0; j < 10; j++) {
    if (j < s[i] - '0') {
      ans += solve(i + 1, cnt + !!j, 1);
    } else if (f || s[i] - '0' == j) {
      ans += solve(i + 1, cnt + !!j, f);
    }
  }
  dp[i][cnt][f] = ans;
  return dp[i][cnt][f];
}
int main() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    long long int l, r;
    cin >> l >> r;
    l--;
    s = to_string(r);
    long long int ans = 0;
    memset(dp, -1, sizeof(dp));
    ans += solve(0, 0, 0);
    memset(dp, -1, sizeof(dp));
    s = to_string(l);
    ans -= solve(0, 0, 0);
    cout << ans << '\n';
  }
}

```