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;
string st, p;
pair<long long, int> b[2005];
long long Mi[2005][2005], n;
long long c[2005];
int dd[2005];
long long Ma;
int main() {
  cin >> st;
  cin >> p;
  memset(b, 30, sizeof(b));
  for (int i = 0; i < st.size(); i++) {
    b[i].second = st.size();
    int u = 0;
    for (int j = i; j <= st.size() - 1; j++) {
      if (st[j] == p[u]) u++;
      if (u == p.size()) {
        b[i].first = j - i + 1 - p.size();
        b[i].second = j;
        break;
      }
    }
  }
  memset(Mi, 30, sizeof(Mi));
  for (int i = st.size() - 1; i >= 0; i--)
    Mi[1][i] = min(Mi[1][i + 1], b[i].first);
  for (int i = 2; i <= st.size(); i++)
    for (int j = st.size() - 1; j >= 0; j--)
      Mi[i][j] = min(Mi[i][j + 1], b[j].first + Mi[i - 1][b[j].second + 1]);
  int u = 0;
  for (int i = 1; i <= st.size(); i++) {
    if (Mi[i][0] == Mi[0][0]) break;
    c[i] = Mi[i][0];
    Ma = max(Ma, c[i]);
    u = i;
  }
  int dem = 0, n = st.size(), m = p.size();
  for (int i = 0; i <= st.size(); i++) {
    if (i <= Ma) {
      cout << upper_bound(c + 1, c + u + 1, 1ll * i) - c - 1 << " ";
      continue;
    }
    while (n - u * m - i < 0) u--;
    cout << u << " ";
  }
}
