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 = 4010;
int len[N];
int ans[N];
int dp[N][N];
int ch[N][26];
char s[N], p[N];
int main() {
  scanf("%s%s", s + 1, p + 1);
  int n = strlen(s + 1);
  int m = strlen(p + 1);
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j <= 25; j++) ch[i][j] = ch[i + 1][j];
    ch[i][s[i] - 'a'] = i;
  }
  for (int i = 1; i <= n; i++) {
    int k = i;
    for (int j = 1; j <= m; j++)
      if (k != 0) {
        k = ch[k][p[j] - 'a'];
        if (k++ == 0) break;
      }
    if (k != 1)
      len[i] = k - i;
    else
      len[i] = n + 1;
  }
  for (int i = 1; i <= n; i++) dp[n + 1][i] = n + 1;
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++) {
      dp[i][j] = min(dp[i + 1][j], len[i] + dp[i + len[i]][j - 1]);
    }
  int k = n;
  while (dp[1][k] == n + 1) k--;
  int j = 0;
  for (int i = 1; i <= k; i++) {
    while (j < dp[1][i] - i * m) ans[j++] = i - 1;
  }
  while (j <= n) ans[j++] = k;
  for (int i = 0; i <= n; i++) ans[i] = min(ans[i], (n - i) / m);
  for (int i = 0; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
