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;
unordered_map<int, int> mp;
const long long mod = (1e9) + 7;
char str1[2005], str2[2005];
int dp[2005][2005];
int len1, len2;
int getd(int x) {
  if (x < len2) return 2005;
  int L1 = x, L2 = len2, ans = 0;
  while (L1 && L2) {
    if (str1[L1] == str2[L2]) {
      L1--;
      L2--;
    } else {
      ans++;
      L1--;
    }
  }
  if (!L2) {
    return ans;
  } else
    return 2005;
}
int main() {
  scanf("%s%s", str1 + 1, str2 + 1);
  len1 = strlen(str1 + 1);
  len2 = strlen(str2 + 1);
  for (int i = 0; i <= len1; i++)
    for (int j = 0; j <= len1; j++)
      if (j > i) dp[i][j] = -3000;
  for (int i = 1; i <= len1; i++) {
    int dln = getd(i);
    for (int j = 0; j <= i; j++) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
    }
    for (int j = 0; j <= i; j++) {
      if (j >= dln) dp[i][j] = max(dp[i][j], dp[i - dln - len2][j - dln] + 1);
    }
  }
  for (int i = 0; i <= len1; i++) {
    printf("%d%c", dp[len1][i], i == len1 ? '\n' : ' ');
  }
  return 0;
}
