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 = 2 * 1000 + 10, M = 5 * 100 + 10, INF = 1e9;
int dp[N][N], m[N], r[N], n, p;
string s, t;
int main() {
  cin >> s >> t;
  n = s.size();
  p = t.size();
  for (int i = 0; i < n; i++) {
    int j, sz = 0, x = 0;
    for (j = i; j < n; j++) {
      if (sz >= t.size()) break;
      if (s[j] == t[sz])
        sz++;
      else
        x++;
    }
    if (sz != t.size())
      m[i] = -1;
    else
      m[i] = x, r[i] = j;
  }
  for (int i = 1; i <= n; i++) dp[n][i] = -INF;
  dp[n][0] = 0;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= n; j++) {
      int cur = -INF;
      if (m[i] >= 0 && j - m[i] >= 0) {
        cur = dp[r[i]][j - m[i]] + 1;
      }
      int lcur = dp[i + 1][j];
      lcur = max(lcur, dp[i + 1][j - 1]);
      dp[i][j] = max(cur, lcur);
    }
  }
  for (int i = 0; i <= n; i++) {
    dp[0][i] = max(0, dp[0][i]);
    cout << dp[0][i] << " ";
  }
  cout << endl;
}
