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;
template <typename TYPE>
inline TYPE sqr(const TYPE &a) {
  return (a) * (a);
}
const int INF = 1000 * 1000 * 1000;
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 2100;
int d[N][N];
int to[N];
int cost[N];
void init(const string &s, const string &p) {
  int n = (int)s.size();
  int m = (int)p.size();
  for (int i = 0; i < int(n); ++i) {
    int k = 0;
    int j;
    int c = 0;
    for (j = i; j < n && k < m; ++j)
      if (s[j] == p[k])
        ++k;
      else if (k)
        ++c;
    if (k >= m) {
      to[i] = j;
      cost[i] = c;
    } else {
      to[i] = -1;
    }
  }
}
int main() {
  string s, p;
  getline(cin, s);
  getline(cin, p);
  init(s, p);
  int n = (int)s.size();
  for (int i = 0; i < int(n); ++i) {
    for (int j = 0; j <= i; ++j) {
      int v = d[i][j];
      d[i + 1][j + 1] = max(d[i + 1][j + 1], v);
      d[i + 1][j] = max(d[i + 1][j], v);
      if (to[i] >= 0) {
        int &dest = d[to[i]][j + cost[i]];
        dest = max(dest, v + 1);
      }
    }
  }
  for (int j = 0; j <= n; ++j) {
    int ans = 0;
    for (int i = 0; i <= n; ++i) ans = max(ans, d[i][j]);
    printf("%d ", ans);
  }
  puts("");
  return 0;
}
