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>
#include<cstring>
#define ll long long
const int L=2e5+5,N=1e5+5;
int n,nk,i,j,now,f[N],g[N][26];
char s[L];
ll cal(ll m){
	if (m<=n) return g[m][now];
	ll l1=n,l2=nk,k,s1=g[l1][now],s2=g[l2][now];
	while (l1+l2<=m)
		k=l1,l1+=l2,l2=k,k=s1,s1+=s2,s2=k;
	return s1+cal(m-l1);
}
int main(){
	scanf("%s",s+1);n=strlen(s+1)/2;
	for (f[0]=-1,i=1,j=-1;i<=n;f[i]=++j,i++)
		for (;j>=0&&s[j+1]!=s[i];j=f[j]);
	nk=n-f[n];
	for (i=1;i<=n;i++){
		for (j=0;j<26;j++) g[i][j]=g[i-1][j];
		g[i][s[i]-97]++;
	}
	ll l,r;scanf("%lld%lld",&l,&r);
	for (now=0;now<26;now++)
		printf("%lld ",cal(r)-cal(l-1));
}