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;
const int inf = 0x3f3f3f3f;
char s[N], p[N];
int ed[N], dp[N][N], ans[N];
int main() {
  cin >> s + 1 >> p + 1;
  int n = strlen(s + 1);
  int m = strlen(p + 1);
  for (int i = 1; i <= n; ++i) {
    int j = 1;
    ed[i] = n + 1;
    for (int k = i; k <= n && j <= m; ++k) {
      if (p[j] == s[k]) ++j;
      if (j == m + 1) ed[i] = k;
    }
  }
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 0; i <= n; ++i) dp[i][0] = 0;
  for (int i = 1; i <= n; ++i) {
    if (ed[i] < n + 1) {
      for (int j = 0; j < n; ++j)
        dp[ed[i]][j + 1] =
            min(dp[i - 1][j] + (ed[i] - i + 1 - m), dp[ed[i]][j + 1]);
    }
    for (int j = 1; j <= n; ++j) dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
  }
  memset(ans, 0, sizeof(ans));
  int maxn = 0;
  for (int i = 0; dp[n][i] != inf; ++i) {
    ans[dp[n][i]] = i;
    maxn = i;
  }
  ans[0] = 0;
  for (int i = 1; i <= n; ++i) {
    int j;
    for (j = 1; j <= m && p[j] == s[i + j - 1]; ++j)
      ;
    if (j == m + 1) ++ans[0], i += m - 1;
  }
  for (int i = 1; i <= n - maxn * m; ++i) ans[i] = max(ans[i], ans[i - 1]);
  for (int i = n - maxn * m + 1; i <= n; ++i) ans[i] = (n - i) / m;
  for (int i = 0; i <= n; ++i) printf("%d ", ans[i]);
  puts("");
  return 0;
}
