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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>

using namespace std;

const int MaxN = 2e5 + 5;

char S[MaxN];
long long Pred[MaxN][26], L, R, T;
int Next[MaxN], n;

void InitPred()
{
	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j < 26; j++) Pred[i][j] = Pred[i - 1][j];
		Pred[i][S[i] - 'a']++;
	}
}

void Kmp()
{
	Next[1] = 0;
	for (int j = 0, i = 2; i <= n; i++)
	{
		for (; j > 0 && S[j + 1] != S[i]; j = Next[j]);
		if (S[j + 1] == S[i])
			j++;
		Next[i] = j;
	}
}

void SwapOffset(long long &a, long long &b)
{
	long long tmp = b;
	b += a;
	a = tmp;
}

long long Fib(long long len, int c)
{
	if (len <= n)
		return Pred[len][c];
	if (len <= n * 2)
		return Pred[n][c] + Pred[len - n][c];
	long long f1 = Pred[n][c], f2 = Pred[n][c] + Pred[T][c], l1 = n, l2 = n + T, tmp;
	while (len > l1 + l2)
	{
		SwapOffset(f1, f2);
		SwapOffset(l1, l2);
	}
	return f2 + Fib(len - l2, c);
}

long long Calc(long long len, int c)
{
	if (len <= n)
		return Pred[len][c];
	if (len <= n * 2)
		return Pred[n][c] + Pred[len - n][c];
	if (n % T == 0)
	{
		long long res = 1LL * (len - n) / T * Pred[T][c] + Pred[n][c];
		len -= n, len %= T;
		return res + Pred[len][c];
	}
	else
		return Fib(len, c);
}

int main()
{
	scanf("%s%lld%lld", S + 1, &L, &R);
	n = strlen(S + 1) / 2;
	InitPred();
	Kmp();
	T = n - Next[n];
	for (int i = 0; i < 26; i++)
		printf("%lld%c", Calc(R, i) - Calc(L - 1, i), i == 25 ? '\n' : ' ');
#ifdef GLOBAL_DEBUG
	system("pause");
#endif
	return 0;
}