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 int N = 5000;
string s, p;
int n, m;
int pi[N];
int dp[N][N];
void calcPi() {
  for (int i = 0; i < n; i++) {
    pi[i] = -1;
    int curr = 0;
    for (int j = i; j < n; j++) {
      if (s[j] == p[curr]) {
        curr++;
        if (curr == m) {
          pi[i] = j + 1;
          break;
        }
      }
    }
  }
}
int main() {
  cin >> s >> p;
  n = s.size();
  m = p.size();
  calcPi();
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] != -1) {
        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 (pi[i] != -1) {
          dp[pi[i]][j + pi[i] - i - m] =
              max(dp[pi[i]][j + pi[i] - i - m], dp[i][j] + 1);
        }
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    printf("%d ", dp[n][j]);
  }
  return 0;
}
