Problem: 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).

Examples

Input

abaaba
6 10


Output

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


Input

xx
1 1000000000000000000


Output

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


Input

vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000


Output

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 CPP):

Solution:
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 200005;
typedef long long i64;

char S[MAX_N];
i64 g[105][26], len[MAX_N];

int N, T;

void work(i64 x, i64 *cnt) {
  memset(g, 0, sizeof g);
  
  int mx = 1;
  for (int i = 1; i <= N; ++i) g[1][S[i] - 'a']++;
  for (int i = 1; i <= T; ++i) g[0][S[i] - 'a']++;
  len[0] = T, len[1] = N;

  while (len[mx] < x) {
    ++mx;
    for (int i = 0; i < 26; ++i) g[mx][i] = g[mx - 1][i] + g[mx - 2][i];
    len[mx] = len[mx - 1] + len[mx - 2];
  }
  
  if (N % T == 0) {
    if (x <= N) {
      for (int i = 1; i <= x; ++i) cnt[S[i] - 'a']++;
    } else {
      for (int i = 1; i <= N; ++i) cnt[S[i] - 'a']++;
      x -= N;
      for (int i = 0; i < 26; ++i) cnt[i] += g[0][i] * (x / T);
      for (int i = 1; i <= x % T; ++i) cnt[S[i] - 'a']++;
    }
  } else {
    while (mx > 1 && x > 0) {
      if (x >= len[mx - 1]) {
	for (int i = 0; i < 26; ++i) cnt[i] += g[mx - 1][i];
	x -= len[mx - 1], mx = mx - 2;
      } else {
	mx = mx - 1;
      }
    }
    for (int i = 1; i <= x; ++i) cnt[S[i] - 'a']++;
  }
}

int main() {
  scanf("%s", S + 1), N = strlen(S + 1);
  N /= 2;
  
  static int f[MAX_N];
  for (int i = 2, j = 0; i <= N; ++i) {
    while (j && S[j + 1] != S[i]) j = f[j];
    if (S[j + 1] == S[i]) j++;
    f[i] = j;
  }

  T = N - f[N];
  i64 l, r;
  scanf("%lld%lld", &l, &r);

  static i64 cnt[2][26];
  work(r, cnt[0]), work(l - 1, cnt[1]);
  for (int i = 0; i < 26; ++i)
    printf("%lld ", cnt[0][i] - cnt[1][i]);
  puts("");
  return 0;
}