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>
const int MOD = 1000000007;
const int INF = 0x3F3F3F3F;
const double EPS = 1e-6;
const int MaxN = 2005;
const int MaxM = 505;
int dp[MaxN][MaxN], n, m;
char a[MaxN], b[MaxM];
int offset[MaxN];
void init() {
  n = strlen(a);
  m = strlen(b);
  for (int i = 0; i < n; i++) {
    int p = i - 1;
    for (int j = 0; p < n && j < m; j++) {
      p = std::find(a + p + 1, a + n, b[j]) - a;
    }
    offset[i] = (p == n) ? -1 : (p - i + 1);
  }
}
int main() {
  while (scanf("%s%s", a, b) == 2) {
    init();
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= i; j++) {
        dp[i + 1][j] = std::max(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = std::max(dp[i + 1][j + 1], dp[i][j]);
        if (offset[i] != -1) {
          dp[i + offset[i]][j + offset[i] - m] =
              std::max(dp[i + offset[i]][j + offset[i] - m], dp[i][j] + 1);
        }
      }
    }
    for (int i = 0; i <= n; i++) {
      printf("%d%c", dp[n][i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}
