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 EPS = -1e8;
const double Pi = acos(-1);
bool inline equ(double a, double b) { return fabs(a - b) < EPS; }
const int MAXN = 2010;
const int MAXP = 510;
int n, p;
char str[MAXN], pat[MAXP];
int prv[MAXN];
int dp[MAXN][MAXN];
int main() {
  scanf("%s%s", str + 1, pat + 1);
  n = strlen(str + 1), p = strlen(pat + 1);
  for (int i = 1; i <= n; i++) {
    int j = i, cnt = p;
    bool flag = 1;
    for (; cnt > 0; cnt--) {
      while (cnt <= j && str[j] != pat[cnt]) j--;
      if (str[j] != pat[cnt] || cnt > j) {
        flag = 0;
        break;
      }
      j--;
    }
    if (flag)
      prv[i] = j;
    else
      prv[i] = -1;
  }
  for (int x = 0; x <= n; x++)
    for (int i = 1; i <= n; i++) {
      dp[x][i] = dp[x][i - 1];
      if (prv[i] == -1) continue;
      if (x >= i) continue;
      int use = i - prv[i] - p;
      if (use > x) continue;
      if (x - use > prv[i]) continue;
      dp[x][i] = max(dp[x][i], dp[x - use][prv[i]] + 1);
    }
  for (int x = 0; x <= n; x++) printf("%d%c", dp[x][n], x == n ? '\n' : ' ');
}
