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>
const int N = 2003;
char s[N], p[N];
int a[N];
int dp[N][N];
bool vst[N][N];
int n, m;
int f(int p, int x) {
  if (x < 0) return -100000008;
  if (vst[p][x]) return dp[p][x];
  vst[p][x] = 1;
  if (p == n) {
    if (x == 0)
      return dp[p][x] = 0;
    else
      return dp[p][x] = -100000008;
  }
  dp[p][x] = f(p + 1, x - 1);
  f(p + 1, x);
  if (dp[p][x] < dp[p + 1][x]) dp[p][x] = dp[p + 1][x];
  if (a[p] != -1) {
    int y = (a[p] - p + 1 - m);
    int v = f(a[p] + 1, x - y) + 1;
    if (dp[p][x] < v) dp[p][x] = v;
  }
  return dp[p][x];
}
int main() {
  int i, j, k;
  scanf("%s%s", &s, &p);
  for (i = 0; s[i]; i++)
    ;
  n = i;
  for (i = 0; p[i]; i++)
    ;
  m = i;
  for (i = 0; s[i]; i++) {
    k = 0;
    for (j = i; s[j]; j++) {
      if (s[j] == p[k]) k++;
      if (!p[k]) break;
    }
    if (s[j])
      a[i] = j;
    else
      a[i] = -1;
  }
  for (i = 0; s[i]; i++) {
    printf("%d ", f(0, i));
  }
  puts("0");
  return 0;
}
