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;
int nxt[2010];
char txt[2010], pat[2010];
int txtsz, patsz;
void pre_process() {
  memset(nxt, -1, sizeof nxt);
  for (int i = 0; i < txtsz; i++) {
    int k = 0, cnt = 0;
    for (int j = 0; i + j < txtsz && k < patsz; j++) {
      if (txt[i + j] == pat[k])
        k++;
      else
        cnt++;
    }
    if (k == patsz) nxt[i] = cnt;
  }
  return;
}
int dp[2010][2010];
int solve(int pos, int left) {
  int &ret = dp[pos][left];
  if (~ret) return ret;
  if (pos == txtsz) {
    if (!left) return ret = 0;
    return ret = -1e9;
  }
  if (left < 0) return ret = -1e9;
  ret = max(solve(pos + 1, left), solve(pos + 1, left - 1));
  if (nxt[pos] != -1)
    ret = max(ret, 1 + solve(pos + nxt[pos] + patsz, left - nxt[pos]));
  return ret;
}
int main() {
  while (scanf("%s %s", txt, pat) == 2) {
    txtsz = strlen(txt);
    patsz = strlen(pat);
    pre_process();
    memset(dp, -1, sizeof dp);
    for (int i = 0; i <= txtsz; i++) printf("%d ", solve(0, i));
    puts("");
  }
  return 0;
}
