#include <bits/stdc++.h>
using namespace std;
char s[2100], p[510];
int lens, lenp, A[2100], dp[2100][2100];
int main() {
  scanf("%s %s", &s, &p);
  lens = strlen(s);
  lenp = strlen(p);
  for (int i = 0; i < lens; i++) {
    int current_index = 0, counter = 0;
    for (int j = i; j < lens; j++) {
      counter++;
      if (s[j] == p[current_index]) current_index++;
      if (current_index == lenp) break;
    }
    if (current_index == lenp)
      A[i] = counter;
    else
      A[i] = -1;
  }
  for (int i = 0; i < lens; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (A[i] >= 0)
        dp[i + A[i]][j + A[i] - lenp] =
            max(dp[i][j] + 1, dp[i + A[i]][j + A[i] - lenp]);
    }
  }
  printf("%d", dp[lens][0]);
  for (int i = 1; i <= lens; i++) printf(" %d", dp[lens][i]);
  printf("\n");
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates <image> that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define <image> as maximum value of <image> over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know <image> for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the <image> for all x from 0 to |s|.

Examples

Input

aaaaa
aa


Output

2 2 1 1 0 0


Input

axbaxxb
ab


Output

0 1 1 2 1 1 0 0

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}. 

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char s[2100], p[510];
int lens, lenp, A[2100], dp[2100][2100];
int main() {
  scanf("%s %s", &s, &p);
  lens = strlen(s);
  lenp = strlen(p);
  for (int i = 0; i < lens; i++) {
    int current_index = 0, counter = 0;
    for (int j = i; j < lens; j++) {
      counter++;
      if (s[j] == p[current_index]) current_index++;
      if (current_index == lenp) break;
    }
    if (current_index == lenp)
      A[i] = counter;
    else
      A[i] = -1;
  }
  for (int i = 0; i < lens; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (A[i] >= 0)
        dp[i + A[i]][j + A[i] - lenp] =
            max(dp[i][j] + 1, dp[i + A[i]][j + A[i] - lenp]);
    }
  }
  printf("%d", dp[lens][0]);
  for (int i = 1; i <= lens; i++) printf(" %d", dp[lens][i]);
  printf("\n");
  return 0;
}

```