#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
string s;
long long dp[20][2][20];
long long solve(long long in, bool c, long long cnt) {
  if (in == s.size()) {
    if (cnt <= 3)
      return 1;
    else
      return 0;
  }
  if (dp[in][c][cnt] != -1) return dp[in][c][cnt];
  long long mx;
  if (c)
    mx = s[in] - '0';
  else
    mx = 9;
  long long ans = 0;
  for (long long i = 0; i <= mx; i++) {
    if (i != 0) {
      ans += solve(in + 1, c & (i == s[in] - '0'), cnt + 1);
    } else {
      ans += solve(in + 1, c & (i == s[in] - '0'), cnt);
    }
  }
  return dp[in][c][cnt] = ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    l--;
    s = to_string(r);
    memset(dp, -1, sizeof(dp));
    long long ans1 = solve(0, 1, 0);
    s = to_string(l);
    memset(dp, -1, sizeof(dp));
    long long ans2 = solve(0, 1, 0);
    cout << ans1 - ans2 << "\n";
  }
}

### Prompt

Your task is to create 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 long long mod = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
string s;
long long dp[20][2][20];
long long solve(long long in, bool c, long long cnt) {
  if (in == s.size()) {
    if (cnt <= 3)
      return 1;
    else
      return 0;
  }
  if (dp[in][c][cnt] != -1) return dp[in][c][cnt];
  long long mx;
  if (c)
    mx = s[in] - '0';
  else
    mx = 9;
  long long ans = 0;
  for (long long i = 0; i <= mx; i++) {
    if (i != 0) {
      ans += solve(in + 1, c & (i == s[in] - '0'), cnt + 1);
    } else {
      ans += solve(in + 1, c & (i == s[in] - '0'), cnt);
    }
  }
  return dp[in][c][cnt] = ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    l--;
    s = to_string(r);
    memset(dp, -1, sizeof(dp));
    long long ans1 = solve(0, 1, 0);
    s = to_string(l);
    memset(dp, -1, sizeof(dp));
    long long ans2 = solve(0, 1, 0);
    cout << ans1 - ans2 << "\n";
  }
}

```