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;
long long n, m, x, dp[2001][2001];
string s, t;
inline long long read() {
  long long sum = 0, x = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return sum * x;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
inline long long work(long long c) {
  if (c < m) return 2e9 + 1e8;
  long long ans = 0, d = m, b = c;
  while (d && b) {
    if (s[b] == t[d]) {
      --d;
      --b;
    } else {
      ++ans;
      --b;
    }
  }
  if (d == 0) return ans;
  return 2e9 + 1e8;
}
signed main() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  s = ' ' + s;
  t = ' ' + t;
  for (register long long i = 0; i <= n; ++i)
    for (register long long j = 0; j <= n; ++j)
      if (i < j) dp[i][j] = -2e9 + 1e8;
  for (register long long i = 1; i <= n; ++i) {
    x = work(i);
    for (register long long j = 0; j <= n; ++j) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j - x >= 0) dp[i][j] = max(dp[i][j], dp[i - m - x][j - x] + 1);
    }
  }
  for (register long long i = 0; i <= n; ++i) {
    write(dp[n][i]);
    putchar(' ');
  }
  return 0;
}
