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 = 1000000000;
const int root = 1;
const int mm = 2222;
char st[mm], p[mm];
int f[mm][mm];
int n, m;
int dp[mm];
int main() {
  scanf("%s%s", st + 1, p + 1);
  n = strlen(st + 1);
  m = strlen(p + 1);
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = m; i <= n; i++)
    if (st[i] == p[m]) {
      int k = i, j;
      for (j = m; j >= 1; j--) {
        while (k >= 1 && st[k] != p[j]) k--;
        if (k < 1) break;
        k--;
      }
      if (j == 0) dp[i] = k;
    }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      f[i][j + 1] = max(f[i][j + 1], f[i - 1][j]);
      f[i][j] = max(f[i][j], f[i - 1][j]);
    }
    if (dp[i] != -1) {
      for (int j = 0; j <= dp[i]; j++)
        f[i][j + i - dp[i] - m] = max(f[i][j + i - dp[i] - m], f[dp[i]][j] + 1);
    }
  }
  for (int i = 0; i <= n; i++) {
    if (i) printf(" ");
    int Max = 0;
    for (int j = 1; j <= n; j++) Max = max(Max, f[j][i]);
    printf("%d", Max);
  }
  printf("\n");
}
