Problem: 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

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#define MAXN 1000010
#define LL long long

int n;
char str[MAXN];
int nxt[MAXN];
int s[26][MAXN];
LL L,R,S,T;

void gaoNext(){
	nxt[1]=0;
	for(int i=2,j=1;i<=n;i++,j++){
		while(j && str[i]!=str[j]){
			if(j==1) j=0;
			else j=nxt[j-1]+1;
		}
		nxt[i]=j;
	}
}

void gaoS(){
	for(int i=0;i<26;i++){
		s[i][0]=0;
		for(int j=1;j<=n;j++)
			s[i][j]=s[i][j-1]+((str[j]=='a'+i)?1:0);
	}
}

void pre_gao(){
	gaoNext();
	int temp=n;
	while(temp*2>=n) temp=nxt[temp];
	n-=temp;
}

LL getS(int ch,LL len){
	if(len<=n) return s[ch][len];
	LL l1=T,l2=S;
	LL s1=s[ch][l1],s2=s[ch][l2];
	while(l1+l2<=len){
		LL l3=l1+l2,s3=s1+s2;
		l1=l2; l2=l3;
		s1=s2; s2=s3;
	}
	return s2+getS(ch,len-l2);
}

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	scanf("%s",str+1);
	n=strlen(str+1);
	scanf("%lld%lld",&L,&R);
	pre_gao();
	gaoS();
	S=n; T=n-nxt[n];
	for(int i=0;i<26;i++)
		printf("%lld ",getS(i,R)-getS(i,L-1));
	return 0;
}