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 MX = 2048;
const int INF = 1029384756;
int N, M;
int rt[MX];
char str[MX], p[MX];
int dp[MX][MX];
vector<pair<int, int>> vec;
int getR(int pos) {
  for (int j = 0; j < M; j++) {
    while (pos < N && p[j] != str[pos]) {
      pos++;
    }
    if (pos >= N) return -1;
    pos++;
  }
  return pos;
}
int go(int i, int j) {
  if (i >= N) {
    if (j == 0) return 0;
    return -INF;
  }
  if (N - i == j) return 0;
  if (j < 0) return -INF;
  if (dp[i][j] != -1) return dp[i][j];
  int res = -INF;
  res = max(res, go(i + 1, j));
  if (rt[i] != -1) {
    res = max(res, go(rt[i], j - (rt[i] - i - M)) + 1);
  }
  return dp[i][j] = res;
}
int main() {
  scanf("%s", str);
  scanf("%s", p);
  N = strlen(str);
  M = strlen(p);
  fill(rt, rt + MX, -1);
  for (int i = 0; i < N; i++) {
    rt[i] = getR(i);
  }
  memset((dp), -1, sizeof(dp));
  for (int i = 0; i <= N; i++)
    printf("%d%c", max(0, go(0, i)), i == N ? '\n' : ' ');
  return 0;
}
