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;
long long qpow(long long a, long long b) {
  long long res = 1, base = a;
  while (b) {
    if (b % 2) res = res * base;
    base = base * base;
    b /= 2;
  }
  return res;
}
long long powmod(long long a, long long b) {
  long long res = 1, base = a;
  while (b) {
    if (b % 2) res = res * base % 1000000007;
    base = base * base % 1000000007;
    b /= 2;
  }
  return res;
}
void scanf(int &__x) {
  __x = 0;
  char __ch = getchar();
  while (__ch == ' ' || __ch == '\n') __ch = getchar();
  while (__ch >= '0' && __ch <= '9')
    __x = __x * 10 + __ch - '0', __ch = getchar();
}
long long gcd(long long _a, long long _b) {
  if (!_b)
    return _a;
  else
    return gcd(_b, _a % _b);
}
int dp[2005][2005];
char s[2005], p[2005];
void init(void) { memset(dp, 0, sizeof dp); }
void read(void) {
  scanf("%s", s + 1);
  scanf("%s", p + 1);
}
void debug(int n) {
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) printf("%d ", dp[i][j]);
    printf("\n");
  }
}
void work(void) {
  int n = strlen(s + 1);
  int m = strlen(p + 1);
  int t, k;
  for (int i = 1; i <= n; i++) {
    for (t = i, k = m; t && k; t--)
      if (s[t] == p[k]) k--;
    if (!k)
      for (int j = 0; j <= i - m; j++)
        if (i - t - m <= j) dp[i][j] = dp[t][j - i + t + m] + 1;
    for (int j = 0; j <= i; j++) {
      if (j) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d%c", dp[n][i], i == n ? '\n' : ' ');
}
int main(void) {
  init();
  read();
  work();
  return 0;
}
