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>
using namespace std;
#define N 200005
#define LL long long
 
char s[N];
int tot[N][26],f[N];
int n,T;
 
LL calc(LL len,int ch)
{
	if(len<=n) return tot[len][ch];
	LL num1=tot[n][ch],num2=tot[T][ch];
	LL len1=n,len2=T;
	while(len1+len2<=len){
		num1=num1+num2,num2=num1-num2;
		len1=len1+len2,len2=len1-len2;
	}
	return num1+calc(len-len1,ch);
}
 
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	f[0]=0;f[1]=0;tot[1][s[1]-'a']=1;
	for(int i=2,p;i<=n;i++)
	{
		p=f[i-1];
		while(p&&s[p+1]!=s[i]) p=f[p];
		f[i]=s[p+1]!=s[i] ? 0:p+1;
		memcpy(tot[i],tot[i-1],sizeof(tot[i]));
		tot[i][s[i]-'a']++;
	}
	T=n-f[n];// cout << T << endl; return 0;
	LL l,r;
	cin>>l>>r;
	for(int i=0;i<26;i++)	cout<<calc(r,i)-calc(l-1,i)<<" ";
	return 0;
}