#include <bits/stdc++.h>
using namespace std;
long long C[20][20];
int pw[5];
void pre() {
  C[0][0] = 1;
  for (int i = 1; i <= 20; ++i) {
    C[i][0] = C[i - 1][0];
    for (int j = 1; j <= i; ++j) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
  pw[0] = 1;
  for (int i = 1; i < 5; ++i) {
    pw[i] = pw[i - 1] * 9;
  }
}
string toS(long long x) {
  string ret;
  while (x) {
    ret += ((x % 10) + '0');
    x /= 10;
  }
  reverse((ret).begin(), (ret).end());
  return ret;
}
long long calc(string s, int lim) {
  if (lim < 0) {
    return 0ll;
  }
  if (lim == 0) {
    return 1ll;
  }
  if (s.empty() || s == "0") {
    return 1ll;
  }
  long long ret = 0;
  for (int i = '0'; i < s[0]; ++i) {
    for (int j = 0; j + (i != '0') <= lim; ++j) {
      ret += C[s.size() - 1][j] * pw[j];
    }
  }
  ret += calc(s.substr(1, s.size() - 1), lim - (s[0] != '0'));
  return ret;
}
void solve() {
  long long l, r;
  scanf("%lld %lld", &l, &r);
  printf("%lld\n", calc(toS(r), 3) - calc(toS(l - 1), 3));
}
int main() {
  int tt;
  scanf("%d", &tt);
  pre();
  while (tt--) {
    solve();
  }
  return 0;
}

### 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 C[20][20];
int pw[5];
void pre() {
  C[0][0] = 1;
  for (int i = 1; i <= 20; ++i) {
    C[i][0] = C[i - 1][0];
    for (int j = 1; j <= i; ++j) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
  pw[0] = 1;
  for (int i = 1; i < 5; ++i) {
    pw[i] = pw[i - 1] * 9;
  }
}
string toS(long long x) {
  string ret;
  while (x) {
    ret += ((x % 10) + '0');
    x /= 10;
  }
  reverse((ret).begin(), (ret).end());
  return ret;
}
long long calc(string s, int lim) {
  if (lim < 0) {
    return 0ll;
  }
  if (lim == 0) {
    return 1ll;
  }
  if (s.empty() || s == "0") {
    return 1ll;
  }
  long long ret = 0;
  for (int i = '0'; i < s[0]; ++i) {
    for (int j = 0; j + (i != '0') <= lim; ++j) {
      ret += C[s.size() - 1][j] * pw[j];
    }
  }
  ret += calc(s.substr(1, s.size() - 1), lim - (s[0] != '0'));
  return ret;
}
void solve() {
  long long l, r;
  scanf("%lld %lld", &l, &r);
  printf("%lld\n", calc(toS(r), 3) - calc(toS(l - 1), 3));
}
int main() {
  int tt;
  scanf("%d", &tt);
  pre();
  while (tt--) {
    solve();
  }
  return 0;
}

```