#include <bits/stdc++.h>
using namespace std;
int stop[2005];
char s[2005], p[505];
int dp[2005][2005];
int main() {
  scanf("%s%s", s, p);
  int slen = strlen(s);
  for (int i = 0; s[i]; ++i) {
    int j = 0;
    for (stop[i] = i; p[j] && stop[i] < slen; ++stop[i]) {
      if (s[stop[i]] == p[j]) ++j;
    }
    if (p[j]) stop[i] = -1;
  }
  int plen = strlen(p);
  for (int i = 0; s[i]; ++i)
    for (int j = 0; s[j]; ++j) dp[i][j] = -1000000;
  dp[0][0] = 0;
  for (int i = 0; s[i]; ++i) {
    for (int j = 0; s[j]; ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (stop[i] != -1) {
        int newj = stop[i] - i - plen + j;
        dp[stop[i]][newj] = max(dp[stop[i]][newj], dp[i][j] + 1);
      }
    }
  }
  for (int i = 0; i <= slen; ++i) {
    printf("%d", dp[slen][i]);
    if (i < slen)
      printf(" ");
    else
      printf("\n");
  }
}

### Prompt

Your challenge is to write 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;
int stop[2005];
char s[2005], p[505];
int dp[2005][2005];
int main() {
  scanf("%s%s", s, p);
  int slen = strlen(s);
  for (int i = 0; s[i]; ++i) {
    int j = 0;
    for (stop[i] = i; p[j] && stop[i] < slen; ++stop[i]) {
      if (s[stop[i]] == p[j]) ++j;
    }
    if (p[j]) stop[i] = -1;
  }
  int plen = strlen(p);
  for (int i = 0; s[i]; ++i)
    for (int j = 0; s[j]; ++j) dp[i][j] = -1000000;
  dp[0][0] = 0;
  for (int i = 0; s[i]; ++i) {
    for (int j = 0; s[j]; ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (stop[i] != -1) {
        int newj = stop[i] - i - plen + j;
        dp[stop[i]][newj] = max(dp[stop[i]][newj], dp[i][j] + 1);
      }
    }
  }
  for (int i = 0; i <= slen; ++i) {
    printf("%d", dp[slen][i]);
    if (i < slen)
      printf(" ");
    else
      printf("\n");
  }
}

```