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<iostream>
#include<cstring>
using namespace std;

int nx[200005],n;
long long a[200][26],le[200],l,r;
char ch[200005];

void kmp()
{
	int j=0;
	for(int i=2;i<=n/2;i++)
	{
		while(j&&ch[j+1]!=ch[i]) 
			j=nx[j];
		if(ch[j+1]==ch[i]) 
			j++;
		nx[i]=j;
	}
}

long long calc(int c,long long w)
{
	int id=1;
	if(le[id]>=w)
	{
		int s=0;
		for(int i=1;i<=w;i++)
		{
			if(ch[i]-'a'==c) 
				s++;
		}
		return s;
	}
	while(w>=le[id])
	{
		a[id+1][c]=a[id][c]+a[id-1][c];
		le[id+1]=le[id]+le[id-1];
		id++;
	}
	long long s=a[id-1][c];
	if(w==le[id-1]) 
		return s;
	else 
		return s+calc(c,w-le[id-1]);
}

int main()
{
	scanf("%s",ch+1);
	n=strlen(ch+1);
	kmp();
	cin>>l>>r;
	le[0]=n/2-nx[n/2];
	le[1]=n/2;
	for(int i=1;i<=n/2;i++)
	{
		a[1][ch[i]-'a']++;
		if(i<=le[0]) 
			a[0][ch[i]-'a']++;
	}
	for(int i=0;i<=25;i++) 
		printf("%lld ",calc(i,r)-calc(i,l-1));
}