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 M = 1000000007;
string s, p;
int dp[2001][2001];
int can[2001];
int Next[2001][26];
int calc(int i, int k) {
  if (i == s.size()) {
    if (k) return -(1 << 29);
    return 0;
  }
  if (dp[i][k] == -1) {
    dp[i][k] = calc(i + 1, k);
    if (k) dp[i][k] = max(dp[i][k], calc(i + 1, k - 1));
    if (can[i] != -1 && can[i] - i - (signed)p.size() <= k)
      dp[i][k] =
          max(dp[i][k], 1 + calc(can[i], k - (can[i] - i - (signed)p.size())));
  }
  return dp[i][k];
}
int main() {
  cin >> s >> p;
  memset(Next, -1, sizeof(Next));
  for (int i = s.size() - 1; i >= 0; --i) {
    for (int j = 0; j < 26; ++j) Next[i][j] = Next[i + 1][j];
    ;
    Next[i][s[i] - 'a'] = i;
  }
  memset(can, -1, sizeof(can));
  for (int i = 0; i + p.size() <= s.size(); ++i) {
    int cur = i;
    for (int j = 0; j < p.size(); ++j) {
      if (Next[cur][p[j] - 'a'] == -1) {
        cur = -2;
        break;
      }
      cur = Next[cur][p[j] - 'a'] + 1;
    }
    if (cur != -2) can[i] = cur;
  }
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i <= s.size(); ++i) printf("%d ", calc(0, i));
  return 0;
}
