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 INF = (1 << 30);
int ls, lp;
char s[2005], p[505];
int dp[2005][2005];
int get_del_num(int x) {
  if (x < lp) return INF;
  int up = x, down = lp;
  int num = 0;
  while (up > 0 && down > 0) {
    if (s[up] == p[down]) {
      up--;
      down--;
    } else {
      up--;
      num++;
    }
  }
  if (down == 0)
    return num;
  else
    return INF;
}
int main() {
  while (scanf("%s", s + 1) != EOF) {
    scanf("%s", p + 1);
    ls = strlen(s + 1);
    lp = strlen(p + 1);
    for (int i = 0; i <= ls; i++)
      for (int j = 0; j <= ls; j++)
        if (j > i) dp[i][j] = -INF;
    for (int i = 1; i <= ls; i++) {
      int cnt = get_del_num(i);
      for (int j = 0; j <= ls; j++) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        if (cnt <= j) {
          dp[i][j] = max(dp[i][j], dp[i - cnt - lp][j - cnt] + 1);
        }
      }
    }
    for (int i = 0; i <= ls; i++) printf("%d ", dp[ls][i]);
    printf("\n");
    memset(s, 0, sizeof(s));
    memset(p, 0, sizeof(p));
    memset(dp, 0, sizeof(dp));
  }
  return 0;
}
