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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;

#define fir first
#define sec second

typedef long long LL;
typedef pair<int,int> PP;

const int Maxn=2E5+5;

int N,M;
char s[Maxn];
int nxt[Maxn];

LL L,R;
LL len[1005];
LL cnt[1005][30];
LL Ans[30];

void Get_Nxt(){
	int X=1,Y=0;
	while(X<=N){
		if(Y==0 || s[Y]==s[X])
			nxt[++X]=++Y;
		else
			Y=nxt[Y];
	}
}

void Work_1(){
	for(int i=1;i<=M;++i)
		Ans[s[i]-'a']+=R/M;
	for(int i=1;i<=R%M;++i)
		++Ans[s[i]-'a'];
	for(int i=1;i<=M;++i)
		Ans[s[i]-'a']-=(L-1)/M;
	for(int i=1;i<=(L-1)%M;++i)
		--Ans[s[i]-'a'];
}

void Calc(LL X,int flag){
	if(X<=N){
		for(int i=1;i<=X;++i)
			Ans[s[i]-'a']+=flag;
		return ;
	}
	int i=1;
	while(len[i+1]<=X) ++i;
	for(int j=0;j<26;++j)
		Ans[j]+=cnt[i][j]*flag;
	Calc(X-len[i],flag);
}

void Work_2(){
	for(int i=1;i<=M;++i)
		++cnt[0][s[i]-'a'];
	for(int i=1;i<=N;++i)
		++cnt[1][s[i]-'a'];
	len[0]=M,len[1]=N;
	for(int i=2;;++i){
		len[i]=len[i-1]+len[i-2];
		if(len[i]>R) break ;
		for(int j=0;j<26;++j)
			cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
	}
	Calc(R,1),Calc(L-1,-1);
}

int main(){
	scanf("%s",s+1),N=strlen(s+1);
	scanf("%lld %lld",&L,&R); Get_Nxt();
	N>>=1; M=nxt[N+1]-1; M=N-M;
	if(N%M==0) Work_1();
	else Work_2();
	for(int i=0;i<26;++i)
		printf("%lld ",Ans[i]);
	return 0;
}