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 double pi = acos(-1.0);
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 2000 + 11;
vector<int> G[N];
int n, m;
string a, b;
void work() {}
int pa[N];
bool f[N][N];
int d2[N][N];
int main() {
  cin >> a >> b;
  n = a.length();
  m = b.length();
  memset(pa, -1, sizeof(pa));
  memset(f, 0, sizeof(f));
  int i, j;
  for (i = 0; i < n; i++) {
    int l = i;
    for (j = 0; j < m; j++) {
      while (l < n && a[l] != b[j]) {
        l++;
      }
      if (l == n) break;
      if (j + 1 == m) {
        for (int r = l; r < n; r++) {
          f[i][r] = 1;
          pa[r] = max(pa[r], i);
        }
      }
      l++;
    }
  }
  memset(d2, -1, sizeof(d2));
  for (i = 0; i < n; i++) {
    if (i) {
      for (j = 0; j < n; j++) {
        d2[i][j + 1] = d2[i - 1][j];
        d2[i][j] = max(d2[i][j], d2[i - 1][j]);
      }
    } else {
      d2[0][0] = 0;
      d2[0][1] = 0;
    }
    if (pa[i] == -1) continue;
    int cs = i - pa[i] + 1 - m;
    if (pa[i] == 0) {
      d2[i][cs] = max(d2[i][cs], 1);
    } else {
      for (j = 0; j < n; j++) {
        if (j >= cs) {
          if (d2[pa[i] - 1][j - cs] != -1)
            d2[i][j] = max(d2[i][j], 1 + d2[pa[i] - 1][j - cs]);
        }
      }
    }
  }
  for (i = 0; i < n + 1; i++) {
    if (d2[n - 1][i] != -1)
      printf("%d ", d2[n - 1][i]);
    else
      printf("0 ");
  }
  return 0;
}
