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>
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 2e3 + 5;
const double eps = 1e-9;
using namespace std;
char s[MAXN], p[MAXN];
int a[MAXN], dp[MAXN][MAXN];
int main() {
  scanf("%s %s", s, p);
  int n = strlen(s), m = strlen(p);
  for (int i = 0; i < n; i++) {
    if (i > 0 && a[i - 1] == -1) {
      a[i] = -1;
      continue;
    }
    int j = 0, k = i;
    while (k < n && j < m) {
      if (s[k] == p[j]) j++;
      k++;
    }
    if (j != m)
      a[i] = -1;
    else
      a[i] = k - i;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] >= 0 && i + a[i] <= n)
        dp[i + a[i]][j + a[i] - m] =
            max(dp[i + a[i]][j + a[i] - m], dp[i][j] + 1);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", dp[n][i]);
  printf("\n");
  return 0;
}
