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;
string s, p;
char buf[2005];
int dp_next[2005][505];
int next(int poss, int posp = 0) {
  if (dp_next[poss][posp] != 0) return dp_next[poss][posp];
  if (!posp && s[poss] != p[posp]) return -1;
  if (posp == p.size()) return poss;
  if (poss == s.size()) return -1;
  if (s[poss] == p[posp])
    dp_next[poss][posp] = next(poss + 1, posp + 1);
  else
    dp_next[poss][posp] = next(poss + 1, posp);
  return dp_next[poss][posp];
}
int dp[2005][2005];
bool dp_[2005][2005];
int solve(int pos, int x) {
  if (dp_[pos][x]) return dp[pos][x];
  if (pos == s.size() && x > 0) return -2000000000;
  if (pos == s.size()) return 0;
  int ret = solve(pos + 1, x);
  if (x > 0) ret = max(ret, solve(pos + 1, x - 1));
  if (next(pos) != -1) {
    int to_remove = next(pos) - pos - p.size();
    if (x >= to_remove) ret = max(ret, 1 + solve(next(pos), x - to_remove));
  }
  dp[pos][x] = ret;
  dp_[pos][x] = true;
  return ret;
}
int main() {
  scanf(" %s", buf);
  s = (string)buf;
  scanf(" %s", buf);
  p = (string)buf;
  for (int i = 0; i <= s.size(); i++) {
    if (i > 0) printf(" ");
    printf("%d", solve(0, i));
  }
  return 0;
}
