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 maxn = 2e3 + 5;
const int inf = 0x3f3f3f3f;
char s[maxn], t[maxn];
int lens, lent;
int dp[maxn][maxn];
int calc(int x) {
  if (x < lent) return inf;
  int len = lent, cnt = 0;
  while (len && x) {
    if (s[x] == t[len])
      x--, len--;
    else
      cnt++, x--;
  }
  if (len == 0)
    return cnt;
  else
    return inf;
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  lens = strlen(s + 1);
  lent = strlen(t + 1);
  for (int i = 1; i <= lens; ++i) {
    int del = calc(i);
    for (int j = 0; j <= i; ++j) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (i - lent - del < j - del || j - del < 0) continue;
      if (j - del >= 0)
        dp[i][j] = max(dp[i][j], dp[i - lent - del][j - del] + 1);
    }
  }
  for (int i = 0; i <= lens; ++i) printf("%d ", dp[lens][i]);
  return 0;
}
