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 = 2010;
char s[maxn], t[maxn];
int d[maxn][maxn], n, m, inf = 1e8;
pair<int, int> gao(int i) {
  int p = 1, pre = i;
  for (; i <= n; i++)
    if (s[i] == t[p]) {
      if (p == 1) pre = i;
      if (p == m) return make_pair(i - pre + 1, i);
      p++;
    }
  return make_pair(inf, n + 1);
}
int main() {
  int ans = 0;
  cin >> s + 1;
  cin >> t + 1;
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 0; i <= n; i++)
    for (int j = 1; j <= n; j++) d[i][j] = inf;
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = gao(i);
    int v = tmp.first;
    int k = tmp.second;
    for (int j = 1; j <= n; j++) d[i][j] = min(d[i - 1][j], d[i][j]);
    for (int j = 1; j <= n; j++)
      d[k][j] = min(d[i - 1][j], d[i - 1][j - 1] + v);
  }
  for (int i = n; i >= 0; i--) {
    int ans = 0;
    for (int j = 1; j <= n; j++) {
      if (d[n][j] == j * m && j * m <= i) ans = j;
      if (j * m <= i && j * m + n - d[n][j] >= i) ans = j;
    }
    cout << ans << " ";
  }
}
