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 = 2e3 + 5, maxp = 500 + 5, inf = 1e9;
int lens, lenp;
int pos[maxn];
int f[maxn][maxn];
char s[maxn], p[maxp];
int getint() {
  char ch;
  int ret = 0, p;
  while (!isdigit(ch = getchar()) && ch != '-')
    ;
  p = ch == '-' ? ch = getchar(), -1 : 1;
  while (isdigit(ch)) ret = ret * 10 + ch - '0', ch = getchar();
  return ret * p;
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", p + 1);
  lens = strlen(s + 1);
  lenp = strlen(p + 1);
  for (int i = 1; i <= lens; ++i) {
    pos[i] = -1;
    if (s[i] != p[lenp]) continue;
    int cur = lenp;
    for (int j = i; j >= 1; --j) {
      if (s[j] == p[cur]) cur--;
      if (!cur) {
        pos[i] = j;
        break;
      }
    }
  }
  for (int i = 1; i <= lens; ++i) {
    if (i < lenp) continue;
    f[i][0] = f[i - 1][0];
    bool flag = true;
    for (int j = 1; j <= lenp; ++j)
      if (s[i - j + 1] != p[lenp - j + 1]) {
        flag = false;
        break;
      }
    f[i][0] = max(f[i][0], f[i - lenp][0] + flag);
  }
  for (int i = 1; i <= lens; ++i) f[0][i] = -inf;
  for (int i = 1; i <= lens; ++i) {
    int del = i - pos[i] + 1 - lenp;
    for (int j = 1; j <= i - lenp; ++j) {
      f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]);
      if ((~pos[i]) && j - del >= 0)
        f[i][j] = max(f[i][j], f[pos[i] - 1][j - del] + 1);
    }
  }
  for (int i = 0; i <= lens - 1; ++i) printf("%d ", f[lens][i]);
  printf("%d\n", f[lens][lens]);
  return 0;
}
