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 = 2000;
string s, p;
int ns, np, leastTill[N + 1], dp[N + 1][N + 1];
int main(int argc, char const *argv[]) {
  cin >> s >> p;
  ns = s.size();
  np = p.size();
  for (int i = 0; i < ns; i++) {
    int ptr = 0, till = -1;
    for (int j = i; j < ns; j++) {
      if (p[ptr] == s[j]) {
        ptr++;
        if (ptr == np) {
          till = j;
          break;
        }
      }
    }
    leastTill[i] = till;
  }
  dp[0][0] = (np == 1 && p[0] == s[0] ? 1 : 0);
  dp[0][1] = 0;
  if (leastTill[0] != -1) {
    int del = (leastTill[0] + 1) - np;
    dp[leastTill[0]][del] = max(dp[leastTill[0]][del], 1);
  }
  for (int i = 0; i < ns - 1; i++) {
    for (int j = 0; j <= i + 1; j++) {
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (leastTill[i + 1] != -1) {
        int del = (leastTill[i + 1] - (i + 1) + 1) - np;
        dp[leastTill[i + 1]][j + del] =
            max(dp[leastTill[i + 1]][j + del], 1 + dp[i][j]);
      }
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
    }
  }
  for (int j = 0; j <= ns; j++) {
    printf("%d ", dp[ns - 1][j]);
  }
  return 0;
}
