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 sgn(double x) { return (x > 1e-8) - (x < -1e-8); }
int count_bit(int x) { return x == 0 ? 0 : count_bit(x >> 1) + (x & 1); }
template <class T>
inline void ckmin(T &a, const T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T &a, const T b) {
  if (b > a) a = b;
}
const int MAXS = 2000 + 10;
const int MAXT = 500 + 10;
char S[MAXS], T[MAXT];
int minlen[MAXS], dp[MAXS][MAXS];
void update(int &x, int y) { x = max(x, y); }
int main() {
  while (scanf("%s%s", S, T) != EOF) {
    int lens = strlen(S), lent = strlen(T);
    for (int i = 0; i < lens; ++i) {
      int pi = i;
      bool fail = false;
      for (int j = 0; j < lent; ++j) {
        while (pi < lens && S[pi] != T[j]) {
          ++pi;
        }
        if (pi == lens) {
          fail = true;
        }
        ++pi;
      }
      if (!fail) {
        minlen[i] = pi - i;
      } else {
        minlen[i] = -1;
      }
    }
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < lens; ++i) {
      for (int j = 0; j <= i; ++j) {
        update(dp[i + 1][j], dp[i][j]);
        update(dp[i + 1][j + 1], dp[i][j]);
        if (minlen[i] != -1) {
          update(dp[i + minlen[i]][j + minlen[i] - lent], dp[i][j] + 1);
        }
      }
    }
    for (int i = 0; i <= lens; ++i) {
      printf("%d%c", dp[lens][i], " \n"[i == lens]);
    }
  }
  return 0;
}
