#include <bits/stdc++.h>
using namespace std;
int dig[22];
void fd(long long n) {
  for (int i = 0; i <= 19; i++) {
    dig[i] = n % 10;
    n /= 10;
  }
}
long long dp[20][3][11][5];
long long call(int pos, int big, int prv, int nz) {
  if (pos < 0) return 1;
  if (nz == 3) return 1;
  if (dp[pos][big][prv][nz] != -1) return dp[pos][big][prv][nz];
  long long ans = 0;
  int hi = 10;
  if (big) {
    hi = dig[pos];
  }
  int j = 0;
  for (int i = 0; i < hi; i++) {
    ans += call(pos - 1, 0, i, nz + j);
    if (i == 0) j++;
  }
  if (hi != 10) ans += call(pos - 1, 1, hi, nz + j);
  return dp[pos][big][prv][nz] = ans;
}
int main() {
  long long n, m, T;
  cin >> T;
  while (T--) {
    cin >> m >> n;
    memset(dp, -1, sizeof(dp));
    fd(n);
    long long ans = call(18, 1, 0, 0);
    memset(dp, -1, sizeof(dp));
    fd(m - 1);
    ans -= call(18, 1, 0, 0);
    cout << ans << endl;
  }
  return 0;
}

### 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;
int dig[22];
void fd(long long n) {
  for (int i = 0; i <= 19; i++) {
    dig[i] = n % 10;
    n /= 10;
  }
}
long long dp[20][3][11][5];
long long call(int pos, int big, int prv, int nz) {
  if (pos < 0) return 1;
  if (nz == 3) return 1;
  if (dp[pos][big][prv][nz] != -1) return dp[pos][big][prv][nz];
  long long ans = 0;
  int hi = 10;
  if (big) {
    hi = dig[pos];
  }
  int j = 0;
  for (int i = 0; i < hi; i++) {
    ans += call(pos - 1, 0, i, nz + j);
    if (i == 0) j++;
  }
  if (hi != 10) ans += call(pos - 1, 1, hi, nz + j);
  return dp[pos][big][prv][nz] = ans;
}
int main() {
  long long n, m, T;
  cin >> T;
  while (T--) {
    cin >> m >> n;
    memset(dp, -1, sizeof(dp));
    fd(n);
    long long ans = call(18, 1, 0, 0);
    memset(dp, -1, sizeof(dp));
    fd(m - 1);
    ans -= call(18, 1, 0, 0);
    cout << ans << endl;
  }
  return 0;
}

```