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>
typedef long long ll;
char s[200010];
int n,nxt[200010],m,cnt[100010][26],k;
ll pos[110],cnts[110][26];
ll calc(ll x,int c){
	if(n%m){
		ll sum=0;
		for(int i=k;i--;)
			if(x>=pos[i])sum+=cnts[i][c],x-=pos[i];
		return sum+cnt[x][c];
	}
	else return cnt[m][c]*(x/m)+cnt[x%m][c];
}
int main(){
	ll l,r;
	scanf("%s%lld%lld",s,&l,&r);
	for(n=0;s[n];n++){
		int&i=nxt[n+1];
		for(i=nxt[n];i&&s[i]!=s[n];i=nxt[i]);
		if(n&&s[i]==s[n])i++;
	}
	n/=2;
	for(int i=0;i<n;i++)
		for(int j=0;j<26;j++)
			cnt[i+1][j]=cnt[i][j]+(s[i]==j+'a');
	m=n-nxt[n];
	if(n%m){
		pos[0]=n,pos[k=1]=n+m;
		for(;pos[k]<=r;k++)pos[k+1]=pos[k-1]+pos[k];
		for(int i=0;i<k;i++)
			for(int j=0;j<26;j++)cnts[i][j]=!i?cnt[n][j]:i==1?cnt[n][j]+cnt[m][j]:cnts[i-2][j]+cnts[i-1][j];
	}
	for(int i=0;i<26;i++)
		printf("%lld%c",calc(r,i)-calc(l-1,i)," \n"[i==25]);
}