#include <bits/stdc++.h>
using namespace std;
string s;
int sz;
int dp[30][30][2];
int f(int pos, int cnt, int flag) {
  if (pos == sz) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (dp[pos][cnt][flag] != -1) return dp[pos][cnt][flag];
  int ans = 0;
  int lim = 9;
  if (flag) lim = (int)(s[pos] - '0');
  for (int i = 0; i <= lim; ++i) {
    if (i == lim and flag)
      ans += f(pos + 1, cnt + (i != 0), 1);
    else
      ans += f(pos + 1, cnt + (i != 0), 0);
  }
  return dp[pos][cnt][flag] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    memset(dp, -1, sizeof dp);
    long long x;
    cin >> x;
    x--;
    s = to_string(x);
    sz = s.size();
    int ans1 = f(0, 0, 1);
    memset(dp, -1, sizeof dp);
    cin >> s;
    sz = s.size();
    int ans2 = f(0, 0, 1);
    cout << ans2 - ans1 << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
string s;
int sz;
int dp[30][30][2];
int f(int pos, int cnt, int flag) {
  if (pos == sz) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (dp[pos][cnt][flag] != -1) return dp[pos][cnt][flag];
  int ans = 0;
  int lim = 9;
  if (flag) lim = (int)(s[pos] - '0');
  for (int i = 0; i <= lim; ++i) {
    if (i == lim and flag)
      ans += f(pos + 1, cnt + (i != 0), 1);
    else
      ans += f(pos + 1, cnt + (i != 0), 0);
  }
  return dp[pos][cnt][flag] = ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    memset(dp, -1, sizeof dp);
    long long x;
    cin >> x;
    x--;
    s = to_string(x);
    sz = s.size();
    int ans1 = f(0, 0, 1);
    memset(dp, -1, sizeof dp);
    cin >> s;
    sz = s.size();
    int ans2 = f(0, 0, 1);
    cout << ans2 - ans1 << "\n";
  }
  return 0;
}

```