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 = 2005;
const int INF = 0x3f3f3f3f;
int n, m, a[maxn], b[maxn], dp[maxn][maxn];
char s[maxn], p[maxn];
void preProcess() {
  int i, j, k;
  memset(dp, -1, sizeof(dp));
  for (i = 0; i < n; i++) {
    b[i] = 0;
    for (k = 0, j = i; k < m && j < n; j++) {
      if (p[k] == s[j])
        k++;
      else
        b[i]++;
    }
    if (k != m) {
      a[i] = b[i] = -1;
    } else {
      a[i] = j - i;
    }
  }
  a[n] = b[n] = -1;
}
int dfs(int i, int x) {
  if (i == n) return x ? -INF : 0;
  if (dp[i][x] != -1) return dp[i][x];
  int res = dfs(i + 1, x);
  if (x > 0) res = max(res, dfs(i + 1, x - 1));
  if (a[i] != -1 && x >= b[i]) res = max(res, dfs(i + a[i], x - b[i]) + 1);
  return dp[i][x] = res;
}
int main() {
  int i, j;
  scanf("%s%s", s, p);
  n = strlen(s);
  m = strlen(p);
  preProcess();
  for (i = 0; i <= n; i++) {
    if (i) putchar(' ');
    int x = dfs(0, i);
    printf("%d", x < 0 ? 0 : x);
  }
  putchar('\n');
  return 0;
}
