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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
string s, p;
int a[2105];
int dfs(int i) {
  while (i < n && s[i] != p[0]) i++;
  if (i == n) return -1;
  int st = i;
  i++;
  int j = 1;
  while (j < m) {
    while (i < n && s[i] != p[j]) i++;
    if (i == n) return -1;
    j++;
    i++;
  }
  return i - 1;
}
int dp[2105][2105];
int main() {
  cin >> s >> p;
  n = s.size();
  m = p.size();
  for (int i = 0; i < n; i++) a[i] = dfs(i);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] != -1) {
        dp[a[i] + 1][j + (a[i] - i + 1 - m)] =
            max(dp[i][j] + 1, dp[a[i] + 1][j + (a[i] - i + 1 - m)]);
      }
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", dp[n][i]);
  return 0;
}
