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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
vector<int>Z(char *s)
{
	vector<int>z(1000010);
	int l=0,r=0;
	int ln=strlen(s);
	for(int k=2;k<=ln;k++)
	{
		if(s[k-1]!=s[0])continue;
		if(k>r)
		{
			l=r=k;
			while(r<=ln && s[r-1]==s[r-l])r++;
			r--;
			z[k]=r+1-l;
		}
		else
		{
			int tk=k+1-l;
			if(z[tk]<r+1-k)
			{
				z[k]=z[tk];
			}
			else
			{
				l=k;
				while(r<=ln && s[r-1]==s[r-l])r++;
				r--;
				z[k]=r+1-l;
			}
		}
	}
	return z;
}
char s[200010];
long long l,r;
int p;
int n;
long long sum[200010][26];
long long gtans(long long x,int y)
{
	if(x<=n)return sum[x][y];
	if(x<=2*n)return sum[x-n][y]+sum[n][y];
	long long s1=sum[n][y]+sum[p][y],s2=sum[n][y],a=n+p,b=n;
	while(x>a+b)
	{
		s1+=s2;s2=s1-s2;a+=b;b=a-b;
	}
	return s1+gtans(x-a,y);
}
long long solve(long long x,int y)
{
	if(x<=n)return sum[x][y];
	if(n%p==0)
	{
		long long t=(x-n)/(long long)p*sum[p][y]+sum[n][y];
		x%=(long long)p;
		return t+sum[x][y];
	}
	else return gtans(x,y);
}
int main()
{
	scanf("%s",s);
	scanf("%lld%lld",&l,&r);
	n=strlen(s)/2;
	vector<int>res=Z(s);
	for(int i=n+2;i<=strlen(s);i++)
	{
		if(res[i]==strlen(s)+1-i)
		{
			p=res[i];
			break;
		}
	}
	p=n-p;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<26;j++)sum[i][j]+=sum[i-1][j];
		sum[i][s[i-1]-'a']++;
	}
	for(int i=0;i<26;i++)
	{
		printf("%lld ",solve(r,i)-solve(l-1,i));
	}
	return 0;
}