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 long long INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF = 0x7ffffff;
const int maxn = 1e5 + 100;
const int mod = 1000000007;
const double pi = acos(-1);
const double eps = 1e-10;
complex<double> rotate(complex<double> v, double a) {
  return v * polar(1.0, a);
}
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};
char s[2222], p[2222];
int slen, plen;
void init() {
  scanf("%s%s", s + 1, p + 1);
  slen = strlen(s + 1), plen = strlen(p + 1);
}
int dp[2222][2222];
void solve() {
  memset(dp, 0, sizeof(dp));
  int st = -1;
  for (int i = 1, j = 1; i <= slen; i++) {
    if (s[i] == p[j]) j++;
    if (j > plen) {
      st = i;
      break;
    }
  }
  if (st == -1) goto label;
  for (int i = st; i <= slen; i++) {
    for (int k = i, j = plen; k >= 1; k--) {
      if (s[k] == p[j]) j--;
      if (j <= 0) {
        st = k - 1;
        break;
      }
    }
    for (int j = 0; j <= i; j++) {
      if (j < i - st - plen)
        dp[i][j] = max(dp[i - 1][j], dp[st][j]);
      else if (st >= j - (i - st - plen))
        dp[i][j] = max(dp[st][j - (i - st - plen)] + 1, dp[i - 1][j]);
    }
  }
label:;
  for (int i = 0; i <= slen; i++) {
    printf("%d", dp[slen][i]);
    if (i == slen)
      puts("");
    else
      printf(" ");
  }
}
int main() {
  init();
  solve();
  return 0;
}
