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 MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN(2005);
struct pp {
  int f, l, cost;
  bool operator<(const pp &other) const {
    if (l != other.l) return l < other.l;
    return f < other.f;
  }
} d[MAXN];
int nd = 0, dp[2005][2005];
int main() {
  string a, b;
  cin >> a >> b;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] == b[b.size() - 1]) {
      int now = b.size() - 1;
      for (int j = i; j >= 0; j--) {
        if (a[j] == b[now]) {
          now--;
          if (now < 0) {
            d[nd].f = j;
            d[nd].l = i;
            d[nd].cost = (i - j + 1) - (b.size());
            nd++;
            break;
          }
        }
      }
    }
  }
  for (int i = 0; i <= a.size(); i++)
    for (int j = 0; j <= a.size(); j++) dp[i][j] = -INF;
  dp[0][0] = 0;
  sort(d, d + nd);
  int now = 1;
  for (int i = 0; i < nd; i++) {
    int f = d[i].f + 1, l = d[i].l + 1, w = d[i].cost;
    while (now <= l) {
      for (int j = a.size(); j >= 0; j--) {
        if (j > 0) dp[now][j] = max(dp[now][j], dp[now - 1][j - 1]);
        dp[now][j] = max(dp[now][j], dp[now - 1][j]);
      }
      now++;
    }
    for (int j = a.size(); j >= 0; j--) {
      if (j - w < 0) break;
      dp[l][j] = max(dp[l][j], dp[f - 1][j - w] + 1);
    }
  }
  while (now <= a.size()) {
    for (int j = a.size(); j >= 0; j--) {
      if (j > 0) dp[now][j] = max(dp[now][j], dp[now - 1][j - 1]);
      dp[now][j] = max(dp[now][j], dp[now - 1][j]);
    }
    now++;
  }
  for (int i = 0; i <= a.size(); i++) {
    if (dp[a.size()][i] == -INF)
      printf("0 ");
    else
      printf("%d ", dp[a.size()][i]);
  }
}
