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 maxn = 2e3 + 10, oo = maxn;
int n, m;
char s[maxn], t[maxn];
int a[maxn];
int dp[maxn][maxn];
void myMax(int &a, int b) {
  if (a < b) {
    a = b;
  }
}
int main() {
  int i, j, k, x, y;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  for (i = n; i >= m; i--) {
    for (j = i, k = 0; j; j--) {
      if (s[j] == t[m - k] && ++k == m) {
        break;
      }
    }
    a[i] = k == m ? i - j + 1 : oo;
  }
  for (; i; i--) {
    a[i] = oo;
  }
  dp[0][0] = 0;
  for (i = (1); i <= (n); i++) {
    for (j = (0); j <= (i); j++) {
      dp[i][j] = 0;
      if (j < i) {
        dp[i][j] = dp[i - 1][j];
      }
      if (j) {
        myMax(dp[i][j], dp[i - 1][j - 1]);
      }
      if (a[i] != oo && (y = j - a[i] + m) >= 0 && (x = i - a[i]) >= y) {
        myMax(dp[i][j], dp[x][y] + 1);
      }
    }
  }
  for (i = (0); i <= (n); i++) {
    printf("%d%c", dp[n][i], " \n"[i == n]);
  }
  return 0;
}
