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>
#pragma warning(disable : 4996)
using namespace std;
char s1[2000 + 1], s2[500 + 1];
int C[2000], P[2000];
int D[2000 + 1][2000 + 1];
int N, M;
int main() {
  scanf("%s%s", s1, s2);
  N = strlen(s1), M = strlen(s2);
  int i, j;
  for (i = 0; i < N; i++) {
    for (j = 0; j < M; j++) {
      while (i + j + C[i] < N && s1[i + j + C[i]] != s2[j]) C[i]++;
      if (i + j + C[i] == N) break;
      if (j == M - 1) P[i] = i + j + C[i] + 1;
    }
    if (j < M) C[i] = -1, P[i] = -1;
  }
  for (i = 0; i <= N; i++) {
    for (j = 0; j <= N; j++)
      if (N - i < j) D[i][j] = -9999999;
  }
  for (i = N - 1; i >= 0; i--) {
    if (P[i] == -1)
      continue;
    else {
      if (C[i] == 0)
        D[i][0] = D[P[i]][0] + 1;
      else
        D[i][0] = D[i + 1][0];
    }
    for (j = 1; j <= N; j++) {
      if (N - i < j) {
        continue;
      }
      D[i][j] = D[i + 1][j];
      if (D[i][j] < D[i + 1][j - 1]) D[i][j] = D[i + 1][j - 1];
      if (P[i] == -1) continue;
      if (j - C[i] >= 0 && D[i][j] < D[P[i]][j - C[i]] + 1)
        D[i][j] = D[P[i]][j - C[i]] + 1;
    }
  }
  for (i = 0; i <= N; i++) {
    printf("%d ", D[0][i]);
  }
  return 0;
}
