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;
string s;
long long l,r;
int p;
int n;
int f[100005];
long long num[200010][26];
long long doit(long long x,int y)
{
	if(x<=n)return num[x][y];
	if(x<=2*n)return num[n][y]+num[x-n][y];
	long long x1=num[n][y]+num[p][y],x2=num[n][y];
	long long l1=n+p,l2=n;
	while(x>l1+l2)
	{
//		cout<<x1<<" "<<x2<<" "<<l1<<" "<<l2<<endl;
		long long t=x1;
		x1+=x2;
		x2=t;
		t=l1;
		l1+=l2;
		l2=t;
	}
	x1+=doit(x-l1,y);
//	cout<<x<<" "<<x1<<endl;
	return x1;
}
long long sol(long long x,int y)
{
	if(x<=n)return num[x][y];
//	if(x<=2*n)return num[n][y]+num[x-n][y];
	if(n%p==0)
	{
		long long xx=(x-n)/(long long)p*num[p][y]+num[n][y];
		x%=(long long)p;
		xx+=num[x][y];
		return xx;
	}
	return doit(x,y);
}
int main()
{
	cin>>s>>l>>r;
	n=s.size()/2;
	s=s.substr(0,n);
	s=" "+s;
	f[1]=0;
	for(int i=2,j=0;i<=n;i++)
	{
		while(s[i]!=s[j+1]&&j>0)j=f[j];
		j+=(s[i]==s[j+1]);
		f[i]=j;
	}
//	cout<<f[0]<<f[1]<<f[2]<<f[3]<<endl;
	p=n-f[n];
//	cout<<p<<endl;
//	p=n-p;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<26;j++)num[i][j]+=num[i-1][j];
		num[i][s[i]-'a']++;
	}
	for(int i=0;i<26;i++)cout<<sol(r,i)-sol(l-1,i)<<" ";
	cout<<endl;
	return 0;
}