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 dx[] = {0, 0, -1, 1};
int dy[] = {-1, 1, 0, 0};
const int INF = 1000000010;
int n, m;
int dp[2010][2010];
int cost[2010];
char S[2010];
char P[510];
int main() {
  scanf("%s%s", S, P);
  n = strlen(S);
  m = strlen(P);
  for (int i = (0); i < (n); ++i) {
    cost[i] = INF;
    int j = i;
    int p = 0;
    while (p < m && j < n) {
      if (S[j] == P[p]) ++p;
      ++j;
    }
    if (p == m) cost[i] = j - i;
  }
  for (int i = (0); i < (n + 1); ++i)
    for (int j = (0); j < (n + 1); ++j) dp[i][j] = (j == 0 ? 0 : INF);
  for (int i = (n - 1); i > (-1); --i) {
    for (int j = (0); j < (n + 1); ++j) {
      dp[i][j] = dp[i + 1][j];
      if (j && cost[i] < INF)
        dp[i][j] = min(dp[i][j], dp[i + cost[i]][j - 1] + cost[i] - m);
    }
  }
  for (int i = (0); i < (n + 1); ++i) {
    int r = 0;
    while (r < n && dp[0][r + 1] <= i) ++r;
    while (r * m + i > n) --r;
    printf("%d%c", r, "\n "[i < n]);
  }
  return 0;
}
