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 str1[2010], str2[2010];
int n1, n2;
void read() {
  scanf("%s %s", str1 + 1, str2 + 1);
  n1 = strlen(str1 + 1), n2 = strlen(str2 + 1);
}
void print() {
  static int dp[2010][2010], to[2010], cost[2010];
  for (int i = 1; i <= n1; ++i) {
    int j = 1, k = i;
    while (j <= n2 && k <= n1) {
      while (k <= n1 && str1[k] != str2[j]) ++k, ++cost[i];
      if (k <= n1) ++k, ++j;
    }
    if (j <= n2)
      cost[i] = 1 << 30;
    else
      to[i] = k;
  }
  for (int i = n1; i >= 1; --i)
    for (int j = 0; j <= n1 - i + 1; ++j) {
      dp[i][j] = dp[i + 1][j];
      if (j - cost[i] <= n1 - to[i] + 1 && j >= cost[i])
        dp[i][j] = max(dp[i][j], dp[to[i]][j - cost[i]] + 1);
    }
  for (int i = 0; i <= n1; ++i) printf("%d%c", dp[1][i], i < n1 ? ' ' : '\n');
}
int main() {
  read();
  print();
  return 0;
}
