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<bits/stdc++.h>
typedef long long i64;
const int N=2e6+7;
char s[N];
int f[N],n,n1,n2,n3;
i64 l,r;
void ext(){
	for(int i=2,j=0;i<=n;++i){
		while(j&&s[i]!=s[j+1])j=f[j];
		if(s[i]==s[j+1])++j;
		f[i]=j;
	}
	int p=f[n];
	while(p*2>=n)p=f[p];
	memcpy(s+n-p+1,s+1,n-p);
	n+=n-p*2;
}
int t[333],ts[2][333];
i64 ans[333];
void cal1(i64 x,int sgn){
	memset(t,0,sizeof(t));
	for(int i=1;i<=x%n;++i)++t[s[i]];
	for(int i='a';i<='z';++i)ans[i]+=t[i]*sgn;
	for(int i=x%n+1;i<=n;++i)++t[s[i]];
	for(int i='a';i<='z';++i)ans[i]+=t[i]*sgn*(x/n);
}
void cal1(){
	n=n2-n1;
	cal1(r,1),cal1(l-1,-1);
}
i64 fs[207][2],len[207];
int mx;
void cal2(i64 x,int sgn){
	for(int i=mx;i>=2;--i)if(len[i]<=x){
		x-=len[i];
		for(int j='a';j<='z';++j)ans[j]+=sgn*(fs[i][0]*ts[0][j]+fs[i][1]*ts[1][j]);
	}
	cal1(x,sgn);
}
void cal2(){
	n1/=2,n2/=2,n3/=2,n/=2;
	for(int i=1;i<=n1;++i)++ts[0][s[i]];
	for(int i=n1+1;i<=n2;++i)++ts[1][s[i]];
	len[0]=n1,len[1]=n2;
	fs[0][0]=fs[1][0]=fs[1][1]=1;
	mx=1;
	do{
		++mx;
		len[mx]=len[mx-1]+len[mx-2];
		fs[mx][0]=fs[mx-1][0]+fs[mx-2][0];
		fs[mx][1]=fs[mx-1][1]+fs[mx-2][1];
	}while(len[mx]<r);
	cal2(r,1),cal2(l-1,-1);
}
int main(){
	scanf("%s%lld%lld",s+1,&l,&r);
	n=strlen(s+1);
	ext();
	n1=n;
	ext();
	n2=n;
	ext();
	n3=n;
	if(n2-n1==n3-n2)cal1();
	else cal2();
	for(int i='a';i<='z';++i)printf("%lld ",ans[i]);
	return 0;
}