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;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long inl() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
inline double ind() {
  double x;
  scanf("%lf", &x);
  return x;
}
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int power(int x, int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0) {
    int y = power(x, n / 2);
    return y * y;
  } else
    return x * power(x, n - 1);
}
int a[2010], dp[2010][2010];
char s[2010], p[2010];
int main() {
  scanf("%s", s);
  scanf("%s", p);
  int n = strlen(s), m = strlen(p), pos = 0, idx = 0;
  for (int i = 0; i < n; i++) {
    pos = i;
    idx = 0;
    while (pos < n && idx < m) {
      if (s[pos] == p[idx]) idx++;
      pos++;
    }
    if (idx == m)
      a[i] = pos - i;
    else
      a[i] = -1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i] != -1) {
        dp[i + a[i]][j + a[i] - m] =
            max(dp[i + a[i]][j + a[i] - m], 1 + dp[i][j]);
      }
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", dp[n][i]);
  return 0;
}
