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 n, sz;
string a;
string b;
int dp[2007][2007];
int cost[2007][2007];
int f(int st) {
  int l = 1;
  int r = st;
  int mid;
  while (r - l > 3) {
    mid = (l + r) / 2;
    if (cost[mid][st] != -1) {
      l = mid;
    } else {
      r = mid;
    }
  }
  while (r > l && cost[r][st] == -1) {
    r--;
  }
  return r;
}
void input();
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
void input() {
  cin >> a;
  a = ('#' + a);
  n = a.size() - 1;
  cin >> b;
  sz = b.size();
}
void solve() {
  int i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      dp[i][j] = -1000000007;
    }
  }
  for (i = 1; i <= n; i++) {
    int cur = 0;
    int br = 0;
    for (j = i; j <= n; j++) {
      if (cur != sz && a[j] != b[cur]) {
        br++;
      }
      if (cur != sz && a[j] == b[cur]) {
        cur++;
      }
      if (cur == sz) {
        cost[i][j] = br;
      } else {
        cost[i][j] = -1;
      }
    }
  }
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= i; j++) {
      if (a[i] == b[sz - 1]) {
        int id = f(i);
        if (cost[id][i] <= j) {
          if (cost[id][i] >= 0 && dp[id - 1][j - cost[id][i]] >= 0) {
            dp[i][j] = max(dp[i][j], dp[id - 1][j - cost[id][i]] + 1);
          }
        }
      }
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j != 0 && dp[i - 1][j - 1] >= 0) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  for (i = 0; i <= n; i++) {
    printf("%d", dp[n][i]);
    if (i == n) {
      printf("\n");
    } else {
      printf(" ");
    }
  }
}
