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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 2100;
const int inf = 1000000000;
char s[maxN], p[maxN];
int n, m;
int l[maxN];
int d[maxN][maxN];
int main() {
  scanf("%s", &s);
  scanf("%s", &p);
  n = strlen(s);
  m = strlen(p);
  for (int i = 0; i < n; ++i) {
    int cnt = 0;
    l[i] = -1;
    for (int j = i; j >= 0; --j) {
      if (s[j] == p[m - 1 - cnt]) {
        ++cnt;
        if (cnt == m) {
          l[i] = j;
          break;
        }
      }
    }
  }
  for (int i = 0; i < maxN; ++i) {
    for (int j = 0; j < maxN; ++j) {
      d[i][j] = inf;
    }
  }
  d[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      d[i][j] = min(d[i][j], d[i - 1][j]);
      if (l[i - 1] == -1) {
        continue;
      }
      if (j > 0) {
        int added = (i - l[i - 1] - m);
        d[i][j] = min(d[i][j], d[l[i - 1]][j - 1] + added);
      }
    }
  }
  for (int i = 0; i <= n; ++i) {
    int res = 0;
    for (int j = 0; j <= n; ++j) {
      if (d[n][j] == inf) {
        continue;
      }
      int a = d[n][j];
      int b = n - j * m;
      if (i >= a && i <= b) {
        res = j;
      }
    }
    printf("%d ", res);
  }
  printf("\n");
  return 0;
}
