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;
const int MXT = 2e3 + 9, MXP = 5e2 + 9, INF = 1e9 + 7;
char text[MXT], pat[MXP];
int dp[MXT][MXT];
int tLen, pLen;
int removable[MXT];
inline void calcDeletions() {
  for (int i = 0; i < tLen; ++i) {
    int ptr1 = i, ptr2 = 0;
    while (ptr1 < tLen && ptr2 < pLen) {
      if (text[ptr1] == pat[ptr2])
        ++ptr1, ++ptr2;
      else
        ++ptr1, ++removable[i];
    }
    if (ptr2 != pLen) removable[i] = INF;
  }
}
inline int go(int idx, int Del) {
  if (Del < 0) return -INF;
  if (idx >= tLen) return ((Del == 0) ? (0) : (-INF));
  if (dp[idx][Del] != -1) return dp[idx][Del];
  int ans = -INF;
  ans = max(ans, max(go(idx + 1, Del - 1), go(idx + 1, Del)));
  ans = max(ans, 1 + go(idx + pLen + removable[idx], Del - removable[idx]));
  return dp[idx][Del] = ans;
}
int main() {
  scanf("%s %s", text, pat);
  tLen = strlen(text), pLen = strlen(pat);
  calcDeletions();
  memset(dp, -1, sizeof(dp));
  for (int Del = 0; Del <= tLen; ++Del) printf("%d ", go(0, Del));
  return 0;
}
