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 = 2002;
int n, m, dp[N][N], cost[N];
char str[N], pattern[N];
int solve(int i, int x) {
  if (x < 0) return -1e9;
  if (i >= n) return x ? -1e9 : 0;
  if (dp[i][x] != -1) return dp[i][x];
  dp[i][x] = max(solve(i + 1, x), solve(i + 1, x - 1));
  if (cost[i] != -1)
    dp[i][x] = max(dp[i][x], solve(i + cost[i] + m, x - cost[i]) + 1);
  return dp[i][x];
}
int main() {
  scanf("%s%s", str, pattern);
  n = (int)strlen(str), m = (int)strlen(pattern);
  memset(cost, -1, sizeof cost);
  for (int i = (0); i <= (int)(n - 1); ++i) {
    int j = 0;
    for (int k = (i); k <= (int)(n - 1); ++k) {
      if (str[k] == pattern[j]) ++j;
      if (j == m) {
        cost[i] = k - i + 1 - m;
        break;
      }
    }
  }
  memset(dp, -1, sizeof dp);
  for (int i = (0); i <= (int)(n); ++i) printf("%d ", solve(0, i));
  printf("\n");
  return 0;
}
