#include <bits/stdc++.h>
const int MAXPOS = 21, MAXCNT = 5;
long long bit[MAXPOS], f[MAXPOS][MAXCNT];
long long dfs(int pos, int cnt, bool lim) {
  if (pos == -1) return 1;
  if (!lim and ~f[pos][cnt]) return f[pos][cnt];
  long long ans = 0, hi = lim ? bit[pos] : 9;
  for (int i = 0; i <= hi; i++) {
    if (!i)
      ans += dfs(pos - 1, cnt, lim and !bit[pos]);
    else if (cnt != 3)
      ans += dfs(pos - 1, cnt + 1, lim and bit[pos] == i);
  }
  if (!lim) f[pos][cnt] = ans;
  return ans;
}
long long solve(long long x) {
  int tot = -1;
  while (x > 0) bit[++tot] = x % 10, x /= 10;
  return dfs(tot, 0, true);
}
int main() {
  memset(f, -1, sizeof(f));
  int T;
  scanf("%d", &T);
  while (T--) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
const int MAXPOS = 21, MAXCNT = 5;
long long bit[MAXPOS], f[MAXPOS][MAXCNT];
long long dfs(int pos, int cnt, bool lim) {
  if (pos == -1) return 1;
  if (!lim and ~f[pos][cnt]) return f[pos][cnt];
  long long ans = 0, hi = lim ? bit[pos] : 9;
  for (int i = 0; i <= hi; i++) {
    if (!i)
      ans += dfs(pos - 1, cnt, lim and !bit[pos]);
    else if (cnt != 3)
      ans += dfs(pos - 1, cnt + 1, lim and bit[pos] == i);
  }
  if (!lim) f[pos][cnt] = ans;
  return ans;
}
long long solve(long long x) {
  int tot = -1;
  while (x > 0) bit[++tot] = x % 10, x /= 10;
  return dfs(tot, 0, true);
}
int main() {
  memset(f, -1, sizeof(f));
  int T;
  scanf("%d", &T);
  while (T--) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    printf("%lld\n", solve(r) - solve(l - 1));
  }
  return 0;
}

```