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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 200000 + 1000;

struct State
{
	long long cnt[26], size;
	State()
	{
		memset(cnt, 0, sizeof(cnt));
		size = 0;
	}
	void add(char c)
	{
		cnt[c - 'a']++, size++;
	}
};

State operator + (const State &a, const State &b)
{
	State ret;
	for(int i = 0; i < 26; i++)
		ret.cnt[i] = a.cnt[i] + b.cnt[i];
	ret.size = a.size + b.size;
	return ret;
}

int n, m;
long long L, R;
char s[MAXN];
State f[MAXN];

State prefix(int v)
{
	State ret;
	for(int i = 1; i <= v; i++)
		ret.add(s[i]);
	return ret;
}

State calc(int x, long long v)
{
	if(x == 0 || x == 1)
		return prefix(v);
	if(v <= f[x - 1].size)
		return calc(x - 1, v);
	else
		return f[x - 1] + calc(x - 2, v - f[x - 1].size);
}

int get_period()
{
	static int next[MAXN];
	for(int i = 2; i <= n; i++)
	{
		int j = next[i - 1];
		while(j && s[j + 1] != s[i])
			j = next[j];
		if(s[j + 1] == s[i])
			j = j + 1;
		next[i] = j;
	}
	return n - next[n];
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> (s + 1);
	cin >> L >> R;
	n = strlen(s + 1) / 2;
	s[n + 1] = 0;

	int k = get_period();
	f[0] = prefix(k);
	f[1] = prefix(n);

	m = 2;
	while(f[m - 1].size <= R) {
		f[m] = f[m - 1] + f[m - 2];
		m++;
	}

	State ans1 = calc(m, L - 1);
	State ans2 = calc(m, R);

	for(int i = 0; i < 26; i++)
		cout << ans2.cnt[i] - ans1.cnt[i] << ' ';
	cout << endl;

	return 0;
}
