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

typedef long long s64;

const int MaxN = 200000;
const int MaxM = MaxN / 2;
const int NLet = 26;

int n;
char s[MaxN + 2];

int sum[MaxM + 1][NLet];

int fail[MaxM + 1];

int cir_l = 0;

inline void build_fail()
{
	int k = 0;
	for (int i = 2; i <= n; ++i)
	{
		while (k && s[k + 1] != s[i])
			k = fail[k];
		fail[i] = (k += s[k + 1] == s[i]);
	}
}

s64 solve(s64 l, int c)
{
	s64 res = 0;
	while (l > n)
	{
		s64 nl = cir_l, nr = n;
		s64 sl = sum[cir_l][c], sr = sum[n][c];

		while (l >= nl + nr)
		{
			nr += nl, nl = nr - nl;
			sr += sl, sl = sr - sl;
		}

		l -= nr, res += sr;
	}

	return res + sum[l][c];
}

int main()
{
	scanf("%s", s + 1);
	n = strlen(s + 1) / 2;
	s[n + 1] = '\0';

	for (int i = 1; i <= n; ++i)
	{
		for (int c = 0; c < NLet; ++c)
			sum[i][c] = sum[i - 1][c];
		++sum[i][s[i] - 'a'];
	}

	build_fail();

	cir_l = n - fail[n];

	s64 l, r;
	cin >> l >> r;

	for (int c = 0; c < NLet; ++c)
		cout << solve(r, c) - solve(l - 1, c) << ' ';
	cout << endl;

	return 0;
}