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 = 2222;
const int INF = 1111111;
char s[N], p[N];
int n, m;
int cost[N][N];
int f[N][N];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", p + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      cost[i][j] = INF;
    }
  for (int i = 1; i <= n + 1; i++) {
    cost[i][m + 1] = 0;
  }
  for (int i = n; i >= 1; i--)
    for (int j = m; j >= 1; j--) {
      cost[i][j] = min(cost[i][j], cost[i + 1][j] + 1);
      if (s[i] == p[j]) {
        cost[i][j] = min(cost[i][j], cost[i + 1][j + 1]);
      }
    }
  for (int i = 0; i <= n + 1; i++)
    for (int j = 0; j <= n + 1; j++) {
      f[i][j] = -INF;
    }
  f[1][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      f[i + 1][j] = max(f[i + 1][j], f[i][j]);
      f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
      if (cost[i][1] != INF && i + cost[i][1] + m <= n + 1 &&
          j + cost[i][1] <= n) {
        f[i + cost[i][1] + m][j + cost[i][1]] =
            max(f[i + cost[i][1] + m][j + cost[i][1]], f[i][j] + 1);
      }
    }
  for (int i = 0; i <= n; i++) {
    printf("%d%c", f[n + 1][i], i == n ? '\n' : ' ');
  }
  return 0;
}
