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 <cstring>
#include <cstdio>

typedef long long LL;
const int N=100005;

int fail[N],n,t;
LL cnt[90][26],f[90],ans[26];
char s[N*2];

void count(int pos,LL *a){
	for (int i=1;i<=pos;i++) a[s[i]-'a']++;
}

void calc(LL pos){
	if (pos<=n){
		count(pos,ans);
		return;
	}
	int m=0;
	for (;f[m+1]<=pos;m++);
	for (int i=0;i<26;i++) ans[i]+=cnt[m][i];
	calc(pos-f[m]);
}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	fail[0]=-1;
	for (int i=1,j=-1;i<=n;fail[i++]=++j)
		for (;j>=0 && s[j+1]!=s[i];j=fail[j]);
	t=n-fail[n];
	f[0]=t;count(t,cnt[0]);
	f[1]=n;count(n,cnt[1]);
	LL l,r;
	scanf("%lld%lld",&l,&r);
	for (int i=2;f[i-1]<=r;i++){
		f[i]=f[i-1]+f[i-2];
		for (int j=0;j<26;j++) cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
	}
	calc(l-1);
	for (int i=0;i<26;i++) ans[i]*=-1;
	calc(r);
	for (int i=0;i<26;i++) printf("%lld ",ans[i]);
}