#include <bits/stdc++.h>
using namespace std;
template <class T>
void _read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  if (f) x = -x;
}
int T;
const int MAXD = 18 + 7;
long long L, R;
int a[MAXD], dp[MAXD][MAXD];
long long dfs(long long pos, bool lim, int cnt) {
  if (cnt > 3) return 0;
  if (pos == -1) return 1;
  if (!lim && dp[pos][cnt] != -1) return dp[pos][cnt];
  long long ret = 0, up = (lim ? a[pos] : 9);
  for (int i = 0; i <= up; i++)
    ret += dfs(pos - 1, lim && (i == up), cnt + (i > 0));
  if (!lim) dp[pos][cnt] = ret;
  return ret;
}
long long solve(long long x) {
  memset(dp, -1, sizeof(dp));
  long long pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, true, 0);
}
int main() {
  _read(T);
  while (T--) {
    _read(L);
    _read(R);
    printf("%lld\n", solve(R) - solve(L - 1));
  }
}

### Prompt

Your challenge is to write 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;
template <class T>
void _read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  if (f) x = -x;
}
int T;
const int MAXD = 18 + 7;
long long L, R;
int a[MAXD], dp[MAXD][MAXD];
long long dfs(long long pos, bool lim, int cnt) {
  if (cnt > 3) return 0;
  if (pos == -1) return 1;
  if (!lim && dp[pos][cnt] != -1) return dp[pos][cnt];
  long long ret = 0, up = (lim ? a[pos] : 9);
  for (int i = 0; i <= up; i++)
    ret += dfs(pos - 1, lim && (i == up), cnt + (i > 0));
  if (!lim) dp[pos][cnt] = ret;
  return ret;
}
long long solve(long long x) {
  memset(dp, -1, sizeof(dp));
  long long pos = 0;
  while (x) {
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, true, 0);
}
int main() {
  _read(T);
  while (T--) {
    _read(L);
    _read(R);
    printf("%lld\n", solve(R) - solve(L - 1));
  }
}

```