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 MAXN = 2050;
int ecnt[MAXN], go[MAXN], dp[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s, p;
  cin >> s >> p;
  int n = (int)s.size(), k = (int)p.size();
  memset(go, -1, sizeof go);
  for (int i = 0; i < n; i++) {
    int cnt = 0, si = i, pi = 0;
    while (si < n && pi < k) {
      if (s[si] == p[pi])
        si++, pi++;
      else
        si++, cnt++;
    }
    if (pi == k) {
      ecnt[i] = cnt;
      go[i] = si;
    }
  }
  for (int i = 0; i < MAXN; i++)
    for (int j = 0; j < MAXN; j++) dp[i][j] = -987654321;
  dp[n][0] = 0;
  for (int m = n - 1; m >= 0; m--) {
    dp[m][0] = dp[m + 1][0];
    for (int x = 1; x <= n - m; x++)
      dp[m][x] = max(dp[m + 1][x], dp[m + 1][x - 1]);
    if (go[m] != -1) {
      for (int x = ecnt[m]; x <= n - m; x++)
        dp[m][x] = max(dp[m][x], dp[go[m]][x - ecnt[m]] + 1);
    }
  }
  for (int x = 0; x <= n; x++) {
    int mx = 0;
    for (int i = 0; i < n; i++) mx = max(mx, dp[i][x]);
    cout << mx << " ";
  }
}
