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>
using namespace std;
typedef long long ll;
int kmp[200010],len;
char s[200010];
ll p,l,r;
ll sum[30][200010];
ll solve(ll L,int n) {
	if(L<=len) return sum[n][L];
	ll s1=sum[n][p],s2=sum[n][len],l1=p,l2=len,x,y;
	while(l1+l2<L){
		x=s2,y=l2;
		s2+=s1,l2+=l1;
		s1=x,l1=y;
	}
	return s2+solve(L-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=kmp[j];    
        if(s[j+1]==s[i]) j++;    
    	kmp[i]=j;
    	//cout<<i<<" "<<kmp[i]<<endl;
    }
	p=len-kmp[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));
}