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<bits/stdc++.h>
#define next nxt
using namespace std;
char st[500010];
int go[500010][26],next[500010],w,len;
long long work(long long r,int ch)
{
	if (r<=len) return go[r][ch];
	long long n1=go[len/2][ch],n2=go[w][ch],l1=len/2,l2=w;
	while (l1+l2<=r)
	{
		long long tmp=l1;l1+=l2;l2=tmp;
		tmp=n1;n1+=n2;n2=tmp;
	}
	return n1+work(r-l1,ch);
}
int main()
{
	scanf("%s\n",st+1);len=strlen(st+1);
	for (int i=1;i<=len;i++)
	{
		for (int j=0;j<26;j++) go[i][j]=go[i-1][j];
		go[i][st[i]-'a']++;
	}
	next[0]=-1;
	for (int i=1;i<=len;i++)
	{
		int t=next[i-1];
		while ((t!=-1)&&(st[t+1]!=st[i])) t=next[t];
		next[i]=t+1;
	}
	w=next[len];
	while (w>=len/2) w=next[w];
	w=len/2-w;
	long long l,r;scanf("%lld%lld",&l,&r);
	for (int i=0;i<26;i++) printf("%lld ",work(r,i)-work(l-1,i));
}