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;
bool is_prime(long long n) {
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
map<long long, long long> fact(long long n) {
  n = abs(n);
  map<long long, long long> ans;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        n /= i;
        ans[i]++;
      }
    }
  }
  if (n > 1) {
    ans[n]++;
  }
  return ans;
}
inline long long getPow(long long a, long long b) {
  long long res = 1ll, tp = a;
  while (b) {
    if (b & 1ll) {
      res *= tp;
    }
    tp *= tp;
    b >>= 1ll;
  }
  return res;
}
long long vec_mult(const pair<long long, long long>& t1,
                   const pair<long long, long long>& t2,
                   const pair<long long, long long>& t3) {
  const long long &x1 = t1.first, y1 = t1.second;
  const long long &x2 = t2.first, y2 = t2.second;
  const long long &x3 = t3.first, y3 = t3.second;
  return abs((x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1));
}
void ok() {
  cout << "yes" << endl;
  exit(0);
}
void no() {
  cout << "no" << endl;
  exit(0);
}
inline long long nxt() {
  long long x;
  cin >> x;
  return x;
}
const long long N = 2e5 + 5, inf = 8e18;
int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  string s1, s2;
  cin >> s1 >> s2;
  int n = s1.size(), m = s2.size();
  vector<long long> mnL(n, -1);
  for (int i = 0; i < n; i++) {
    long long ind = 0;
    for (int j = i; j < n; j++) {
      if (s1[j] == s2[ind]) {
        ind++;
      }
      if (ind == m) {
        mnL[i] = j;
        break;
      }
    }
  }
  long long dp[n + 1][n + 1];
  fill(&dp[0][0], &dp[0][0] + (n + 1) * (n + 1), 0ll);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      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 (mnL[i] == -1) continue;
      long long dlt = mnL[i] - i + 1 - m;
      dp[mnL[i] + 1][j + dlt] = max(dp[mnL[i] + 1][j + dlt], dp[i][j] + 1);
    }
  }
  for (int i = 0; i <= n; i++) {
    cout << dp[n][i] << " ";
  }
  return 0;
}
