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 kMax = 2010;
const int kInf = 1 << 20;
string s, p;
int N, M;
int f[kMax][kMax], g[kMax][2];
vector<int> res;
void solve();
int main() {
  cin >> s >> p;
  N = int((s).size()), M = int((p).size()), s = "#" + s, p = "#" + p;
  solve();
  return 0;
}
void solve() {
  fill(&g[0][0], &g[kMax][0], -1);
  for (int i = 1; i <= N; ++i) {
    int x = i, y = M, c = 0;
    while (x > 0 && y > 0) {
      if (s[x] == p[y])
        --y;
      else if (y < M)
        ++c;
      --x;
    }
    if (y == 0) g[i][0] = x, g[i][1] = c;
  }
  fill(&f[1][0], &f[kMax][0], kInf);
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= N; ++j) {
      f[i][j] = f[i][j - 1];
      if (~g[j][0]) f[i][j] = min(f[i][j], f[i - 1][g[j][0]] + g[j][1]);
    }
  int rcnt = N;
  while (f[rcnt][N] == kInf) --rcnt;
  for (int i = 1; i <= rcnt; ++i)
    if (f[i - 1][N] != f[i][N])
      res.insert(res.end(), f[i][N] - f[i - 1][N], i - 1);
  for (int i = f[rcnt][N]; i <= N - rcnt * M; ++i) res.push_back(rcnt);
  while ((--rcnt) >= 0) res.insert(res.end(), M, rcnt);
  for (int i = 0; i < int((res).size()); ++i)
    cout << res[i] << (i + 1 == int((res).size()) ? "\n" : " ");
}
