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 long long MOD = 1e9 + 7;
const int MAXN = 2000 + 5;
int dp[MAXN][MAXN] = {0};
int previous[MAXN] = {0};
void calculatePrevious(const string &s, const string &p) {
  int sSize = s.length();
  int pSize = p.length();
  for (int i = 0; i < sSize; ++i) {
    previous[i] = -2;
    int currentP = pSize - 1;
    int currentS = i;
    while (currentP >= 0 && currentS >= 0) {
      while (currentS >= 0 && s[currentS] != p[currentP]) {
        currentS--;
      }
      if (currentS >= 0) {
        currentP--;
        currentS--;
      }
    }
    if (currentP == -1) {
      previous[i] = currentS;
    }
  }
}
int main() {
  string s, p;
  cin >> s >> p;
  int sSize = s.length();
  int pSize = p.length();
  calculatePrevious(s, p);
  for (int i = 1; i <= sSize; ++i) {
    for (int j = 0; j < sSize; ++j) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      int prev = previous[i - 1];
      if (prev != -2) {
        int toDelete = (i - 1) - prev - pSize;
        int newDelete = j - toDelete;
        int newIndex = prev + 1;
        if (newDelete >= 0 && newIndex >= newDelete) {
          dp[i][j] = max(dp[i][j], 1 + dp[newIndex][newDelete]);
        }
      }
    }
  }
  for (int i = 0; i < sSize; ++i) cout << dp[sSize][i] << " ";
  cout << 0 << endl;
  return 0;
}
