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>
#define N 200005
#define LL long long
#define ULL unsigned LL
using namespace std;
LL l,r,a[26],b[26],f[95][27],F[95];
int n; char A[N];
ULL H[N],S[N];
ULL HA(int l,int r){ return H[r]-H[l-1]*S[r-l+1];}
void calc(LL r,int v,LL *a)
{
	int i,j;
	if(n%v==0){
		for(i=1;i<=n&&i<=r;i++) a[A[i]-'a']++;
		if(r>=n){
			r-=n;
			for(i=1;i<=v;i++)
				a[A[i]-'a']+=r/v+(r%v>=i);
		  }
	  }
	else{
		memset(f,0,sizeof(f));
		for(i=1;i<=v;i++) f[0][A[i]-'a']++;
		for(i=1;i<=n;i++) f[1][A[i]-'a']++;
		F[0]=v,F[1]=n;
		for(i=2;i<=90;i++){
			F[i]=F[i-1]+F[i-2];
			for(j=0;j<26;j++)
				f[i][j]=f[i-1][j]+f[i-2][j];
			if(F[i]>r) break;
		  }
		while(i){
			if(r>=F[i]){
				for(j=0;j<26;j++) a[j]+=f[i][j];
				r-=F[i];
			  }
			i--;
		  }
		for(i=1;i<=r;i++) a[A[i]-'a']++;
	  }
}
int main()
{
	int i;
	scanf("%s",A+1);
	n=strlen(A+1)/2;
	cin>>l>>r;
	S[0]=1;
	for(i=1;i<=n;i++)
		H[i]=H[i-1]*173+A[i],S[i]=S[i-1]*173;
	for(i=1;i<=n;i++)
		if(HA(i+1,n)==HA(1,n-i))
			break;
	calc(r,i,a);
	calc(l-1,i,b);
	for(i=0;i<26;i++)
		printf("%lld ",a[i]-b[i]);
	return 0;
}