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 GCC optimize("unroll-loops")
#pragma GCC optimize("O3", "omit-frame-pointer", "inline")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2")
using namespace std;
const long long mod = 1000000007;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int mxN = 2010;
const int mxP = 510;
int dp[mxN][mxN];
int nxt[mxN][mxP];
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  srand(time(NULL));
  cout << fixed << setprecision(9);
  int t = 1, C = 1;
  while (t--) {
    string s, p;
    cin >> s >> p;
    for (int j = 0; j <= (int)p.size(); j++) {
      nxt[(int)s.size()][j] = (int)s.size() + 1;
    }
    nxt[(int)s.size()][(int)p.size()] = (int)s.size();
    for (int i = (int)s.size() - 1; i >= 0; i--) {
      for (int j = 0; j < (int)p.size(); j++) {
        nxt[i][j] = nxt[i + 1][j];
        if (s[i] == p[j]) nxt[i][j] = min(nxt[i][j], nxt[i + 1][j + 1]);
      }
      nxt[i][(int)p.size()] = i;
    }
    dp[0][0] = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      for (int j = 0; j <= i; j++) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
        int del = nxt[i][0] - i - p.size();
        if (j + del > (int)s.size() || nxt[i][0] > (int)s.size()) continue;
        dp[nxt[i][0]][j + del] = max(dp[nxt[i][0]][j + del], dp[i][j] + 1);
      }
    }
    for (int i = 0; i <= (int)s.size(); i++) {
      cout << dp[(int)s.size()][i] << " \n"[i == (int)s.size()];
    }
  }
  return 0;
}
