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 inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int Pow(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base % mo;
  int tmp = Pow(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
char s1[3000], s2[3000];
int dp[2100][2100], l1, l2, reach[3000];
inline void upd(int &x, int y) { x = max(x, y); }
int main() {
  scanf("%s%s", s1 + 1, s2 + 1);
  l1 = strlen(s1 + 1);
  l2 = strlen(s2 + 1);
  for (int i = (1); i <= (l1); ++i) {
    int cur = 0;
    reach[i] = inf;
    for (int j = (i); j <= (l1); ++j) {
      if (s1[j] == s2[cur + 1]) cur++;
      if (cur == l2) {
        reach[i] = j - i + 1;
        break;
      }
    }
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 0;
  for (int i = (0); i <= (l1 - 1); ++i)
    for (int j = (0); j <= (min(l1, i)); ++j) {
      upd(dp[i + 1][j], dp[i][j]);
      upd(dp[i + 1][j + 1], dp[i][j]);
      if (reach[i + 1] < inf && j + reach[i + 1] - l2 <= l1)
        upd(dp[i + reach[i + 1]][j + reach[i + 1] - l2], dp[i][j] + 1);
    }
  for (int i = (0); i <= (l1); ++i) printf("%d ", dp[l1][i]);
  return 0;
}
