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;
char s[2002], p[505];
int fn[2002];
int dp[2002][2002];
int main() {
  gets(s);
  gets(p);
  int n = strlen(s), m = strlen(p);
  for (int i = 0; i < n; ++i) {
    int ptr = i - 1;
    for (int j = 0; j < m && ptr < n; ++j) {
      ++ptr;
      while (ptr < n && s[ptr] != p[j]) ++ptr;
    }
    fn[i] = ptr;
  }
  memset(dp, -1, sizeof(dp));
  dp[n][0] = 0;
  for (int i = n - 1; i >= 0; --i) {
    for (int x = 0; x <= n; ++x) dp[i][x + 1] = dp[i + 1][x];
    for (int x = 0; x <= n; ++x) dp[i][x] = max(dp[i][x], dp[i + 1][x]);
    for (int j = 0; j <= n - i; ++j) dp[i][j] = max(dp[i][j], 0);
    for (int x = 0; x <= n - i; ++x) {
      int sub = fn[i] - i + 1 - m;
      if (x >= sub && fn[i] != n && dp[fn[i] + 1][x - sub] != -1)
        dp[i][x] = max(dp[i][x], dp[fn[i] + 1][x - sub] + 1);
    }
  }
  for (int x = 0; x <= n; ++x) cout << dp[0][x] << " ";
  cout << endl;
  return 0;
}
