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 MAXS = 2000 + 5;
const int MAXP = 500 + 5;
const int INF = INT_MAX / 2;
vector<int> ans;
char s[MAXS], p[MAXP];
int nx[MAXS];
int lens, lenp;
bool usd[MAXS][MAXS];
int rec[MAXS][MAXS];
int f(int i, int k) {
  if (k == 0) return 0;
  if (i >= lens) return INF;
  if (usd[i][k]) return rec[i][k];
  rec[i][k] = f(i + 1, k);
  if (nx[i] < INF)
    rec[i][k] = min(rec[i][k], f(i + nx[i], k - 1) + nx[i] - lenp);
  usd[i][k] = 1;
  return rec[i][k];
}
int main() {
  scanf("%s%s", s, p);
  lens = strlen(s);
  lenp = strlen(p);
  for (int i = 0; i < lens; ++i) {
    nx[i] = INF;
    if (s[i] == p[0]) {
      int a = i + 1, b = 1;
      while (a < lens && b < lenp) {
        if (s[a] == p[b]) {
          ++a;
          ++b;
        } else
          ++a;
      }
      if (b == lenp) nx[i] = a - i;
    }
  }
  int cnt = 0;
  int k;
  for (k = 1; k <= lens; ++k) {
    int tmp = f(0, k);
    if (tmp >= INF) break;
    while (cnt < tmp) {
      ans.push_back(k - 1);
      ++cnt;
    }
  }
  --k;
  if (k == 0) {
    for (int i = 0; i <= lens; ++i) ans.push_back(0);
  } else {
    for (int i = 0; i < lens + 1 - cnt - (lenp * k); ++i) ans.push_back(k);
    --k;
    while (k >= 0) {
      for (int i = 0; i < lenp; ++i) ans.push_back(k);
      --k;
    }
  }
  printf("%d", ans[0]);
  for (int i = 1; i < ans.size(); ++i) printf(" %d", ans[i]);
  printf("\n");
  return 0;
}
