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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
string T, P;
int n, m;
int dp[2011][2011], pos[2011];
int rec(int p, int del) {
  if (p == n) {
    if (del == 0)
      return 0;
    else
      return -1000000007;
  }
  if (~dp[p][del]) return dp[p][del];
  int ret = -1000000007;
  if (del) ret = rec(p + 1, del - 1);
  ret = max(ret, rec(p + 1, del));
  if (pos[p] != -2) {
    int c = pos[p] - p - m;
    if (c <= del) ret = max(ret, 1 + rec(pos[p], del - c));
  }
  return dp[p][del] = ret;
}
int LCS[2011][2011];
int rec1(int i, int j) {
  if (j == m) return i;
  if (i == n) return -2;
  if (~LCS[i][j]) return LCS[i][j];
  int ret = -2;
  if (T[i] == P[j]) ret = rec1(i + 1, j + 1);
  int ret2 = rec1(i + 1, j);
  if (ret2 == -2) return LCS[i][j] = ret;
  if (ret == -2) return LCS[i][j] = ret2;
  return LCS[i][j] = min(ret, ret2);
}
int main() {
  cin >> T >> P;
  n = T.size();
  m = P.size();
  memset((dp), (-1), sizeof(dp));
  memset((LCS), (-1), sizeof(LCS));
  for (int i = 0; i < n; i++) pos[i] = rec1(i, 0);
  for (int i = 0; i < n + 1; i++) {
    int ans = rec(0, i);
    if (ans < 0)
      cout << 0 << ' ';
    else
      cout << ans << ' ';
  }
  return 0;
}
