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;
char s[2040], p[2040];
int ns, np;
int fail[2040];
void buildFail(char *p) {
  int m = strlen(p);
  int j = fail[0] = -1;
  for (int i = 1; i <= m; i++) {
    while (j >= 0 && p[j] != p[i - 1]) j = fail[j];
    fail[i] = ++j;
  }
}
int us[2040][2040];
int best[2040];
void bfs() {
  deque<pair<pair<int, int>, int> > q;
  memset(us, 0x3f, sizeof(us));
  q.push_front(pair<pair<int, int>, int>(pair<int, int>(0, 0), 0));
  us[0][0] = 0;
  while (q.size()) {
    pair<pair<int, int>, int> w = q.front();
    pair<pair<int, int>, int> nw = w;
    q.pop_front();
    int d = us[w.first.first * np + w.first.second][w.second];
    best[d] = max(best[d], w.first.first);
    if (w.second < ns) {
      int k = w.first.second;
      while (k >= 0 && p[k] != s[w.second]) k = fail[k];
      k++;
      if (k == np) {
        nw.first.first++;
        nw.first.second = 0;
      } else
        nw.first.second = k;
      nw.second++;
      if (us[nw.first.first * np + nw.first.second][nw.second] == 0x3f3f3f3f) {
        us[nw.first.first * np + nw.first.second][nw.second] = d;
        q.push_front(nw);
      }
      nw = w;
      nw.second++;
      if (us[nw.first.first * np + nw.first.second][nw.second] == 0x3f3f3f3f) {
        us[nw.first.first * np + nw.first.second][nw.second] = d + 1;
        q.push_back(nw);
      }
    }
  }
}
int main() {
  while (scanf(" %s %s", s, p) == 2) {
    ns = strlen(s);
    np = strlen(p);
    buildFail(p);
    memset(best, 0, sizeof(best));
    bfs();
    for (int i = 0; i < ns + 1; i++) {
      int res = 0;
      for (int j = 0; j <= i; j++) res = max(res, best[j]);
      while (ns - res * np < i) res--;
      printf("%d ", res);
    }
    puts("");
  }
  return 0;
}
