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 N = 2010;
int nxt[N], dp[N][N];
int ns, np, oo = -1e9;
string s, p;
int calc(int in, int x) {
  if (x < 0) return oo;
  if (in >= ns) {
    return !x ? 0 : oo;
  }
  int &ret = dp[in][x];
  if (ret != -1) {
    return ret;
  }
  ret = calc(in + 1, x);
  if (nxt[in] != -1) {
    int temp = (nxt[in] - in + 1) - np;
    ret = max(ret, calc(nxt[in] + 1, x - temp) + 1);
  }
  ret = max(ret, calc(in + 1, x - 1));
  return ret;
}
int main() {
  cin >> s >> p;
  ns = s.length();
  np = p.length();
  memset(dp, -1, sizeof dp);
  memset(nxt, -1, sizeof nxt);
  for (int i = 0; s[i]; i++) {
    int l = i, r = 0;
    while (l < ns && r < np) {
      if (s[l] == p[r])
        l++, r++;
      else
        l++;
    }
    if (r == np) {
      nxt[i] = l - 1;
    }
  }
  for (int x = 0; x <= ns; x++) {
    printf("%d ", calc(0, x));
  }
  return 0;
}
