#include <bits/stdc++.h>
using namespace std;
const int INF = numeric_limits<int>::max();
using ll = long long;
using ii = pair<int, int>;
using vi = vector<ll>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vvvvi = vector<vvvi>;
using vii = vector<ii>;
using vvii = vector<vii>;
using vb = vector<bool>;
using vvb = vector<vb>;
vvvi dp;
vi a;
ll f(int i, int nz, bool le) {
  int n = a.size();
  if (nz == 4) return 0;
  if (i == n) return 1;
  if (dp[i][nz][le] != -1) return dp[i][nz][le];
  ll ans = 0;
  if (le) {
    if (a[i] == 0) {
      ans += f(i + 1, nz, true);
    } else {
      ans += f(i + 1, nz + 1, true);
      if (a[i] >= 1) ans += (a[i] - 1) * f(i + 1, nz + 1, false);
      ans += f(i + 1, nz, false);
    }
  } else {
    ans += f(i + 1, nz, false);
    ans += 9 * f(i + 1, nz + 1, false);
  }
  return dp[i][nz][le] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    ll l, r;
    cin >> l >> r;
    l--;
    a.clear();
    for (char c : to_string(l)) a.push_back(c - '0');
    dp = vvvi(19, vvi(4, vi(2, -1)));
    ll l_count = f(0, 0, true);
    a.clear();
    for (char c : to_string(r)) a.push_back(c - '0');
    dp = vvvi(19, vvi(4, vi(2, -1)));
    ll r_count = f(0, 0, true);
    cout << r_count - l_count << endl;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 INF = numeric_limits<int>::max();
using ll = long long;
using ii = pair<int, int>;
using vi = vector<ll>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vvvvi = vector<vvvi>;
using vii = vector<ii>;
using vvii = vector<vii>;
using vb = vector<bool>;
using vvb = vector<vb>;
vvvi dp;
vi a;
ll f(int i, int nz, bool le) {
  int n = a.size();
  if (nz == 4) return 0;
  if (i == n) return 1;
  if (dp[i][nz][le] != -1) return dp[i][nz][le];
  ll ans = 0;
  if (le) {
    if (a[i] == 0) {
      ans += f(i + 1, nz, true);
    } else {
      ans += f(i + 1, nz + 1, true);
      if (a[i] >= 1) ans += (a[i] - 1) * f(i + 1, nz + 1, false);
      ans += f(i + 1, nz, false);
    }
  } else {
    ans += f(i + 1, nz, false);
    ans += 9 * f(i + 1, nz + 1, false);
  }
  return dp[i][nz][le] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    ll l, r;
    cin >> l >> r;
    l--;
    a.clear();
    for (char c : to_string(l)) a.push_back(c - '0');
    dp = vvvi(19, vvi(4, vi(2, -1)));
    ll l_count = f(0, 0, true);
    a.clear();
    for (char c : to_string(r)) a.push_back(c - '0');
    dp = vvvi(19, vvi(4, vi(2, -1)));
    ll r_count = f(0, 0, true);
    cout << r_count - l_count << endl;
  }
}

```