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 = 2010;
const int inf = ~0U >> 1;
char s[N], p[N];
int n, m;
int pre[N];
int dp[N][N];
void init() {
  memset(pre, -1, sizeof(pre));
  for (int i = 1; i <= n; ++i) {
    int ii = m, j = i;
    while (j >= 1 && ii >= 1) {
      if (s[j] == p[ii]) ii--;
      j--;
    }
    if (ii == 0) pre[i] = j + 1;
  }
}
void DP() {
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) dp[i][j] = 0;
    if (pre[i] == -1) continue;
    int cost = (i - pre[i] + 1) - m;
    for (int j = 0; j <= i; ++j)
      if (dp[pre[i] - 1][j - cost] != -1 && j - cost >= 0)
        dp[i][j] = max(dp[i - 1][j], dp[pre[i] - 1][j - cost] + 1);
      else
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
  }
  for (int i = 0; i <= n; ++i) printf("%d ", dp[n][i]);
}
int main() {
  scanf("%s %s", s + 1, p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  init();
  DP();
  return 0;
}
