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 <cstring>
#include <cstdio>
typedef long long ll; 
int n, k, nxt[200005]; 
ll ans[35]; 
char str[200005]; 
void work(ll r)
{
	if (r <= n)
	{
		for (int i = 0; i < r; i++)
			ans[str[i] - 'a']++; 
		return; 
	}
	ll cnt_s = 1, cnt_t = 0, lst_s = 0, lst_t = 1; 
	while (cnt_s * n + cnt_t * k < r)
	{
		cnt_s += lst_s; 
		cnt_t += lst_t; 
		lst_s = cnt_s - lst_s; 
		lst_t = cnt_t - lst_t; 
	}
	work(r - lst_s * n - lst_t * k); 
	for (int i = 0; i < n; i++)
		ans[str[i] - 'a'] += lst_s; 
	for (int i = 0; i < k; i++)
		ans[str[i] - 'a'] += lst_t; 
}
int main()
{
	// freopen("ARC077-F.in", "r", stdin); 
	scanf("%s", str); 
	n = strlen(str) / 2; 
	nxt[0] = -1; 
	for (int i = 1, j = -1; i < n; i++)
	{
		while (~j && str[i] != str[j + 1])
			j = nxt[j]; 
		j += str[i] == str[j + 1]; 
		nxt[i] = j; 
	}
	k = n - nxt[n - 1] - 1; 
	ll l, r; 
	scanf("%lld%lld", &l, &r); 
	work(l - 1); 
	for (int i = 0; i < 26; i++)
		ans[i] *= -1; 
	work(r); 
	for (int i = 0; i < 26; i++)
		printf("%lld%c", ans[i], " \n"[i == 25]);
	return 0; 
}