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;
int dp[2020][2020];
int dt[2020][2020];
int g[2020][2020];
char s[2020], p[505];
int l[2020], r[2020];
int ans[2020];
int main() {
  scanf(" %s %s", s, p);
  memset(l, -1, sizeof(l));
  memset(r, -1, sizeof(r));
  int n = strlen(s);
  int m = strlen(p);
  for (int i = 0; i < n; ++i) {
    if (s[i] == p[0]) {
      int t = 1, pos = i;
      for (int j = i + 1; t < m && j < n; ++j) {
        if (s[j] == p[t]) {
          ++t;
          pos = j;
        }
      }
      if (t == m) {
        l[pos] = i;
        r[i] = pos;
      }
    }
  }
  memset(dp, 0x3f, sizeof(dp));
  memset(dt, 0x3f, sizeof(dt));
  memset(g, 0x3f, sizeof(g));
  for (int i = 0; i < n; ++i) {
    dp[i][0] = 0;
    dt[i][0] = 0;
    g[i][0] = 0;
    for (int j = 1; j <= n; ++j) {
      if (r[i] != -1) {
        if (i == 0) {
          if (j == 1) {
            dp[i][j] = r[i] - i + 1 - m;
            dt[r[i]][j] = min(dt[r[i]][j], dp[i][j]);
          }
        } else {
          dp[i][j] = g[i - 1][j - 1] + r[i] - i + 1 - m;
          dt[r[i]][j] = min(dt[r[i]][j], dp[i][j]);
        }
        g[r[i]][j] = min(g[r[i]][j], dt[r[i]][j]);
      }
      if (i > 0) g[i][j] = min(g[i - 1][j], dt[i][j]);
    }
  }
  for (int i = n; i >= 0; --i) {
    for (int j = n; j >= 0; --j) {
      if (j * m <= i && i + g[n - 1][j] <= n) {
        ans[i] = j;
        break;
      }
    }
    printf("%d ", ans[i]);
  }
  puts("");
  return 0;
}
