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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
const int N=200010;
typedef long long ll;
int n,prd,fail[N],k=0;
char a[N];
ll l,r,pre[N][30];
inline ll fib(ll p,int c){
	if(p<=n) return pre[p][c];
	if(p<=n*2) return pre[n][c]+pre[p-n][c];
	ll ret1=pre[n][c]+pre[prd][c],ret2=pre[n][c],a=n+prd,b=n;
	while(p>a+b){
		ll c=ret1; ret1+=ret2; ret2=c;
		c=a; a+=b; b=c;
	}
	return ret1+fib(p-a,c);
}
inline ll cal(ll p,int c){
	if(p<=n) return pre[p][c];
	if(n%prd==0){
		ll ret=1LL*(p-n)/prd*pre[prd][c]+pre[n][c];
		p%=prd;
		return ret+pre[p][c];  
	}
	else return fib(p,c);
}
 
int main(){
	scanf("%s",a+1);
	n=strlen(a+1)/2;
	for (int i=1;i<=n;i++){
		for (int j=0;j<26;j++) pre[i][j]=pre[i-1][j];
    	pre[i][a[i]-'a']++;
	}
	scanf("%lld%lld",&l,&r);
	for(int i=2;i<=n;i++){
		while (k&&a[k+1]!=a[i]) k=fail[k];
		if (a[k+1]==a[i]) k++;
		fail[i]=k;
	}
	prd=n-fail[n];
	for(int i=0;i<26;i++)
		printf("%lld ",cal(r,i)-cal(l-1,i));
	return 0;
}