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
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

typedef long long LL;
const int N = 2e5 + 5;

char s[N];
int fail[N], n, cyc;
LL l, r, add[27], sub[27], tots[27], tott[27];
LL tab[90], cnts[90], cntt[90];

inline void solve1() {
  for (int i = 1; i <= cyc; i ++) add[s[i] - 'a'] ++;
  for (int i = 0; i < 26; i ++) {
    sub[i] = add[i] * (l / cyc);
    add[i] *= r / cyc;
  }
  for (int i = 1; i <= l % cyc; i ++) sub[s[i] - 'a'] ++;
  for (int i = 1; i <= r % cyc; i ++) add[s[i] - 'a'] ++;
}

inline void dfs(LL x, LL *res) {
  if (x <= n) {
    for (int i = 1; i <= x; i ++) res[s[i] - 'a'] ++;
    return;
  }
  int p;
  for (p = 1; tab[p + 1] <= x; p ++);
  for (int i = 0; i < 26; i ++) res[i] += cnts[p] * tots[i] + cntt[p] * tott[i];
  dfs(x - tab[p], res);
}

inline void solve2() {
  tab[0] = cyc; tab[1] = n;
  cnts[1] = 1;
  cntt[0] = 1;
  for (int i = 1; tab[i] <= r; i ++) {
    tab[i + 1] = tab[i] + tab[i - 1];
    cnts[i + 1] = cnts[i] + cnts[i - 1];
    cntt[i + 1] = cntt[i] + cntt[i - 1];
  }
  for (int i = 1; i <= n; i ++) tots[s[i] - 'a'] ++;
  for (int i = 1; i <= cyc; i ++) tott[s[i] - 'a'] ++;
  dfs(l, sub);
  dfs(r, add);
}

int main() {
  scanf("%s", s + 1);
  scanf("%lld%lld", &l, &r);
  l --;
  n = strlen(s + 1) >> 1;
  for (int i = 2, j = 0; i <= n; i ++) {
    while (j && s[j + 1] != s[i]) j = fail[j];
    if (s[i] == s[j + 1]) fail[i] = ++ j;
    else fail[i] = j;
  }
  cyc = n - fail[n];
  if (n % cyc == 0) solve1();
  else solve2();
  LL tot = 0;
  for (int i = 0; i < 26; i ++) {
    printf("%lld ", add[i] - sub[i]);
    tot += add[i] - sub[i];
  }
  cerr << "!!!!!!!!!!!!!" << tot << endl;
  return 0;
}
