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 MAX = 2005, oo = 1e9;
int n, m, dp[MAX][MAX];
char a[MAX], b[MAX];
int main() {
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = -oo;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    int k = i;
    for (int l = m; k > 0; k--) {
      if (a[k] == b[l]) l--;
      if (l == 0) break;
    }
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0 && dp[i - 1][j - 1] > dp[i][j]) dp[i][j] = dp[i - 1][j - 1];
    }
    if (k == 0) continue;
    for (int j = 0; j <= i; j++) {
      int x = j - (i - k + 1 - m);
      if (x < 0 || dp[k - 1][x] == -oo) continue;
      if (dp[k - 1][x] + 1 > dp[i][j]) dp[i][j] = dp[k - 1][x] + 1;
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", dp[n][i]);
  printf("%d\n", dp[n][n]);
  return 0;
}
