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 = 2010;
const int INF = 0x3f3f3f3f;
int r[MAXN];
string s, p;
int n;
int memo[MAXN][MAXN];
int dp(int pos, int q) {
  if (q == 0) return 0;
  if (pos > n) return INF;
  if (memo[pos][q] != -1) return memo[pos][q];
  int sem = dp(pos + 1, q);
  if (r[pos] != INF) {
    int com = dp(r[pos] + 1, q - 1) + r[pos] - pos + 1 - (int)p.size();
    return memo[pos][q] = min(com, sem);
  }
  return memo[pos][q] = sem;
}
int main() {
  memset(memo, -1, sizeof memo);
  memset(r, INF, sizeof r);
  cin >> s >> p;
  n = s.size();
  for (int i = 0; i < s.size(); i++) {
    int cnt = 0;
    int k = -1;
    for (int j = i; j < s.size(); j++) {
      if (cnt < p.size() && s[j] == p[cnt]) cnt++;
      if (cnt == p.size()) {
        k = j;
        break;
      }
    }
    if (k != -1) r[i] = k;
  }
  for (int i = 0; i <= n; i++) {
    int resp = 0;
    for (int j = 1; j <= n / ((int)p.size()); j++) {
      int h = dp(0, j);
      if (i >= h && i <= n - (j * (int)p.size())) resp = j;
    }
    cout << resp << " ";
  }
}
