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", ""}.
#include <bits/stdc++.h>
using namespace std;
char s[2005];
char p[505];
int dp[2005][2005];
int len[2005];
int main() {
  scanf("%s", s);
  scanf("%s", p);
  int len_s = strlen(s);
  int len_p = strlen(p);
  for (int i = 0; i < len_s; i++) {
    int tot = i;
    if (s[i] != p[0]) {
      len[i] = -1;
      continue;
    }
    for (int j = 0; j < len_p; j++) {
      while (tot < len_s && s[tot] != p[j]) tot++;
      tot++;
    }
    if (tot >= len_s + 1)
      len[i] = -1;
    else
      len[i] = tot - 1;
  }
  for (int i = 0; i <= len_s; i++)
    for (int j = 0; j <= len_s; j++) dp[i][j] = 0x3f3f3f3f;
  dp[0][0] = 0;
  int all = len_s / len_p;
  len[len_s] = -1;
  for (int i = 0; i <= len_s; i++) {
    if (i) {
      for (int j = 0; j <= all; j++) dp[i][j] = min(dp[i][j], dp[i - 1][j]);
    }
    if (len[i] != -1) {
      for (int j = 0; j <= all; j++) {
        dp[len[i] + 1][j + 1] =
            min(dp[len[i] + 1][j + 1], dp[i][j] + len[i] - i + 1 - len_p);
      }
    }
  }
  int temp = 1;
  while (dp[len_s][temp] <= 2000 &&
         dp[len_s][temp] - dp[len_s][temp - 1] >= 0 && temp <= all) {
    for (int i = 0; i < dp[len_s][temp] - dp[len_s][temp - 1]; i++)
      printf("%d ", temp - 1);
    temp++;
  }
  temp -= 1;
  int a = len_s - temp * len_p - dp[len_s][temp] + 1;
  while (a > 0 && a--) printf("%d ", temp);
  for (int i = temp * len_p - 1; i >= 0; i--) {
    printf("%d ", i / len_p);
  }
}
