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:
//Written by Zhu Zeqi
//Come on,baby
//Hack,please
#include<cmath>
#include<math.h>
#include<ctype.h>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<complex>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cwchar>
#include<cwctype>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define pii pair<int,int>
#define vi vector<int>
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 1000000000
typedef long long ll;
//orz yht
using namespace std;
const int D=26;
char s[200005];
int p[200005];
ll len[1000],num[1000];
int i,j,k,n,m;
ll l,r;
ll work(ll x,int ch){
	int i,j;
	ll ans=0;
	ch+='a';
	len[1]=n;
	num[1]=0;
	for(i=1;i<=n;i++)
	if(s[i]==ch)
	num[1]++;
	len[2]=n+m;
	num[2]=num[1];
	for(i=1;i<=m;i++)
	if(s[i]==ch)
	num[2]++;
	for(i=2;i;i++){
		if(len[i-1]+len[i]>=x)
		break;
		len[i+1]=len[i-1]+len[i];
		num[i+1]=num[i-1]+num[i];
	}
	for(j=i;j;j--){
		if(x>=len[j]){
			x-=len[j];
			ans+=num[j];
		}
	}
	for(i=1;i<=x;i++)
	if(s[i]==ch)
	ans++;
	return ans;
}
int main(){
	//freopen("input.in","r",stdin);
	//freopen("output.out","w",stdout);
	scanf("%s",s+1);
	n=strlen(s+1);
	n/=2;
	j=0;
	for(i=2;i<=n;i++){
		while(j && s[j+1]!=s[i])
		j=p[j];
		if(s[j+1]==s[i])
		p[i]=++j;
	}
	m=n-p[n];
	cin>>l>>r;
	for(i=0;i<D;i++)
	cout<<work(r,i)-work(l-1,i)<<" ";
	puts("");
	//system("pause");
	return 0;
}