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 <bits/stdc++.h>
#define ll long long
using namespace std;
char s[200005];ll len,slen,tlen,sum[30][200005],nex[200005],l,r;
ll solve(ll now,int n)
{
	if(now<=len) return sum[n][now];
	ll f1=sum[n][slen],f2=sum[n][len],l1=slen,l2=len,tf,tl;
	while(l1+l2<now) tf=f2,tl=l2,f2+=f1,l2+=l1,f1=tf,l1=tl;
	return f2+solve(now-l2,n);
}
int main()
{
	scanf("%s%lld%lld",s+1,&l,&r);len=strlen(s+1)/2;
	for(int i=2,j=0;i<=len;i++) 
	{
		while(j&&s[i]!=s[j+1]) j=nex[j];
		if(s[j+1]==s[i]) nex[i]=++j;
	}
	slen=len-nex[len];
	for(int i=0;i<26;i++) for(int j=1;j<=len;j++)
	{
		sum[i][j]=sum[i][j-1];
		if(s[j]-'a'==i) sum[i][j]++;
	}
	for(int i=0;i<26;i++) printf("%lld ",solve(r,i)-solve(l-1,i));
}