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
/*
F(SS)->G(STST)
这里T是S的最短非完整周期，即T是S的前缀，S是T...T的前缀。
那么只需要考虑S->ST的变换，手玩发现规律得
S->ST->STS->STSST...
f(i+2)=f(i+1)+f(i)
每次取最长的一个前缀f(d)删去以后递归即可，O(logW) 
*/

#include<cstdio>
#include<cstring>

typedef long long ll;
const int N=200005;

char s[N]; ll L,R;
int n,m,fail[N],cnt[26][N];

void init()
{
	for(int j=0; j<26; ++j)
		for(int i=1; i<=n; i++)
			cnt[j][i]=cnt[j][i-1]+(s[i]-'a'==j);
	for(int i=2,j; i<=n; i++)
	{
		for(j=fail[i-1];j&&s[j+1]!=s[i];j=fail[j]);
		fail[i]=j+(s[j+1]==s[i]);
	}
	m=n-fail[n];
}

ll solve(int j,ll a)
{
	if(a<=n)
		return cnt[j][a];
	ll p=n,q=m,x=cnt[j][n],y=cnt[j][m],tmp;
	while(p+q<=a)
	{
		tmp=p;
		p+=q;
		q=tmp;
		tmp=x;
		x+=y;
		y=tmp;
	}
	return x+solve(j,a-p);
}

int main()
{
	scanf("%s%lld%lld",s+1,&L,&R);
	n=strlen(s+1)/2;
	init();
	for(int j=0; j<26; j++)
		printf("%lld ",solve(j,R)-solve(j,L-1));
	return 0;
}
