Problem:
Score : 1100 points

Problem Statement

We will call a string that can be obtained by concatenating two equal strings an even string.
For example, xyzxyz and aaaaaa are even, while ababab and xyzxy are not.

For a non-empty string S, we will define f(S) as the shortest even string that can be obtained by appending one or more characters to the end of S.
For example, f(abaaba)=abaababaab.
It can be shown that f(S) is uniquely determined for a non-empty string S.

You are given an even string S consisting of lowercase English letters.
For each letter in the lowercase English alphabet, find the number of its occurrences from the l-th character through the r-th character of f^{10^{100}} (S).

Here, f^{10^{100}} (S) is the string f(f(f( ... f(S) ... ))) obtained by applying f to S 10^{100} times.

Constraints

2 \leq |S| \leq 2\times 10^5

1 \leq l \leq r \leq 10^{18}

S is an even string consisting of lowercase English letters.

l and r are integers.

Input

Input is given from Standard Input in the following format:

S
l r

Output

Print 26 integers in a line with spaces in between.
The i-th integer should be the number of the occurrences of the i-th letter in the lowercase English alphabet from the l-th character through the r-th character of f^{10^{100}} (S).

Sample Input 1

abaaba
6 10

Sample Output 1

3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Since f(abaaba)=abaababaab, the first ten characters in f^{10^{100}}(S) is also abaababaab. Thus, the sixth through the tenth characters are abaab. In this string, a appears three times, b appears twice and no other letters appear, and thus the output should be 3 and 2 followed by twenty-four 0s.

Sample Input 2

xx
1 1000000000000000000

Sample Output 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0

Sample Input 3

vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000

Sample Output 3

87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <cstdio>
#include <cstring>
#include <valarray>

typedef long long i64;

const int N = 200000 + 10, S = 100;

int n, x, y;
char s[N];

std::valarray<i64> sum[N];

void init() {
  static int next[N];
  next[1] = 0;
  for (int i = 2; i <= n; ++i) {
    int j = next[i - 1];
    while (j && s[j + 1] != s[i]) j = next[j];
    j += (s[j + 1] == s[i]);
    next[i] = j;
  }
  sum[0].resize(26);
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1], ++sum[i][s[i] - 'a'];
  x = n - next[n];
  y = x - next[x];
}

std::valarray<i64> solve(i64 m) {
  if (m <= n) return sum[m];
  if (x % y == 0) {
    m -= x;
    return sum[x] + sum[y] * (m / y) + sum[m % y];
  }
  static i64 fib[S];
  static std::valarray<i64> temp[S];
  temp[0] = sum[fib[0] = x];
  temp[1] = sum[fib[1] = x + y];
  for (int i = 2; i < S; ++i) {
    fib[i] = fib[i - 1] + fib[i - 2];
    temp[i] = temp[i - 1] + temp[i - 2];
  }
  int k = 2;
  while (fib[k] < m) ++k;
  std::valarray<i64> res(26);
  while (k > 2) {
    if (m >= fib[k - 1]) {
      res += temp[k - 1];
      m -= fib[k - 1];
      k -= 2;
    } else {
      --k;
    }
  }
  res += sum[m];
  return res;
}

int main() {
  i64 l, r;
  scanf(" %s%lld%lld", s + 1, &l, &r);
  n = strlen(s + 1);
  init();
  std::valarray<i64> ans = solve(r) - solve(l - 1);
  for (int i = 0; i < 26; ++i) printf("%lld ", ans[i]);
  return 0;
}