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 <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define N 200005

char ch[N];
int p[N];
long long a[30];
int n,len;
long long ans[30][1005];
long long length[N];
long long l,r;

void work(long long x,int flag)
{
	if (x == 0) return ;
	long long tmp = 0;
	if (x <= len)
	{
		for(int i = 1;i <= x;i++)
		{
			a[ch[i] - 'a'] += flag;
		}
		return;
	}
	else
	{
		int t = 1;
		while(length[t + 1] <= x) t++;
		for(int i = 0; i < 26;i++)
		{
			a[i] += ans[i][t] * flag;
		}
		work(x - length[t],flag);
	}
}

int main()
{
	//freopen("test.in","r",stdin);
	int j = 0;
	scanf("%s%lld%lld",ch + 1,&l,&r);
	len = strlen(ch + 1);
	//p[1] = 0;
	for(int i = 2;i <= len; i++)
	{
		while(j > 0 && ch[j + 1] != ch[i]) j = p[j];
		if (ch[j + 1] == ch[i]) j++;
		p[i] = j;
	//	printf("%d ",p[i]);
	}
	//puts("");
	len >>= 1;
	n = len - p[len];
	//printf("%d %d\n",len,n);
	if( len % n == 0)
	{
		for(int i = 1;i <= n;i++) a[ch[i] - 'a'] += (long long)r/n;
		for(int i = 1;i <= r % n; i++) a[ch[i] - 'a']++;
		for(int i = 1;i <= n;i++) a[ch[i] - 'a'] -= (long long)(l - 1)/n;
		for(int i = 1;i <= (l - 1) % n;i++) a[ch[i] - 'a']--;	
	}
	else
	{
		for(int i = 1;i <= n; i++)
		{
			ans[ch[i] - 'a'][0]++;
		}
		for(int i = 1;i <= len; i++)
		{
			ans[ch[i] - 'a'][1]++;
		}
		length[0] = n;
		length[1] = len;
		for(int i = 2;;i++)
		{
			length[i] = length[i - 1] + length[i - 2];
			if(length[i] > r) break;
			for(int j = 0; j < 26; j++)
				ans[j][i] = ans[j][i - 1] + ans[j][i - 2];
		}
		work(r,1);work(l - 1,-1);
	}
	for(int i = 0; i < 26; i++)
	{
		printf("%lld ",a[i]);
	}
	return 0;
}