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;
char S[2100], P[510];
int n, m, dp[2100][2100], UP[2100];
bool can[2100][2100];
int main() {
  ios::sync_with_stdio(false);
  cin >> S >> P;
  n = strlen(S);
  m = strlen(P);
  memset(UP, -1, sizeof(UP));
  for (int i = 0; i < n; i++) {
    int j = 0, k = 0;
    while (i + j < n && k < m) {
      if (S[i + j] == P[k]) {
        k++;
        j++;
      } else
        j++;
    }
    if (k == m) {
      for (j--; i + j < n; j++) {
        can[i][i + j] = true;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j >= 0; j--)
      if (can[j][i]) {
        UP[i] = j;
        break;
      }
  }
  for (int i = 0; i < n; i++) {
    if (UP[i] != -1) {
      dp[i][i + 1 - m] = max(dp[i][i + 1 - m], 1);
      for (int k = 0; k <= i + 1; k++) {
        int st = i - UP[i] + 1 - m;
        if (k >= st) {
          if (UP[i] == 0) {
            if (k == st) dp[i][k] = max(dp[i][k], 1);
          } else {
            int pre = UP[i] - 1;
            if (pre + 1 >= k - st) {
              dp[i][k] = max(dp[i][k], dp[pre][k - st] + 1);
            }
          }
        }
      }
    }
    for (int k = 0; k <= i + 1; k++)
      if (i > 0) dp[i][k] = max(dp[i][k], dp[i - 1][k]);
  }
  for (int i = 0; i <= n; i++) {
    cout << dp[n - 1][i] << " ";
  }
  return 0;
}
