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;
int MOD = 1000 * 1000 * 1000 + 7;
const int maxn = 2 * 1000 + 10;
int preCalc[maxn];
int dp[maxn][maxn];
int n;
int m;
int main() {
  string mother, pattern;
  cin >> mother >> pattern;
  n = ((int)(mother).size());
  m = ((int)(pattern).size());
  for (int i = 0; i < n; i++) {
    int ii = i;
    int j = 0;
    while (j < m) {
      while (ii < n && mother[ii] != pattern[j]) {
        ii++;
      }
      if (ii == n) {
        break;
      }
      ii++;
      j++;
    }
    preCalc[i] = ii;
    if (j != m) {
      preCalc[i] = -1;
    }
  }
  dp[n][0] = 0;
  for (int j = 1; j <= n; j++) {
    dp[n][j] = -123123123;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = dp[i + 1][j];
      if (j - 1 >= 0) dp[i][j] = max(dp[i][j], dp[i + 1][j - 1]);
      if (preCalc[i] != -1 && j - (preCalc[i] - i - m) >= 0)
        dp[i][j] = max(dp[i][j], 1 + dp[preCalc[i]][j - (preCalc[i] - i - m)]);
    }
  }
  for (int cnt = 0; cnt <= n; cnt++) {
    int maxi = -1;
    for (int i = 0; i < n; i++) {
      maxi = max(maxi, dp[i][cnt]);
    }
    cout << maxi << ' ';
  }
  cout << endl;
  return 0;
}
