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<cstring>

#define N 200010
#define ll long long

using namespace std;

struct data
{
	ll len,a[30];
	data()
	{
		len=0;
		memset(a,0,sizeof(a));
	}
}f[100];

data operator + (data a,data b)
{
	data c;
	c.len=a.len+b.len;
	for(int i=0;i<26;i++)
		c.a[i]=a.a[i]+b.a[i];
	return c;
}

data operator - (data a,data b)
{
	data c;
	c.len=a.len-b.len;
	for(int i=0;i<26;i++)
		c.a[i]=a.a[i]-b.a[i];
	return c;
}

int n,maxn;
int cut,kmp[N];
char s[N],ss[N];
ll ql,qr;

data query(int k,ll x)
{
	if(x==f[k].len) return f[k];
	if(!k)
	{
		data c;
		c.len=x;
		for(int i=cut+1;i<=cut+x;i++)
			c.a[s[i]-'a']++;
		return c;
	}
	if(k==1)
	{
		data c;
		c.len=x;
		for(int i=1;i<=x;i++)
			c.a[s[i]-'a']++;
		return c;
	}
	if(x<=f[k-1].len) return query(k-1,x);
	return f[k-1]+query(k-2,x-f[k-1].len);
}

int main()
{
	scanf("%s%lld%lld",s+1,&ql,&qr);
	n=strlen(s+1);
	for(int i=1;i<=n/2;i++)
		ss[n/2-i+1]=s[i];
	for(int i=2,j=0;i<=n/2;i++)
	{
		while(j&&ss[j+1]!=ss[i]) j=kmp[j];
		if(ss[j+1]==ss[i]) j++;
		kmp[i]=j;
	}
	int cut=n/2-kmp[n/2];
	for(int i=cut+1;i<=n/2;i++)
	{
		f[0].len++;
		f[0].a[s[i]-'a']++;
	}
	for(int i=1;i<=cut;i++)
	{
		f[1].len++;
		f[1].a[s[i]-'a']++;
	}
	for(int i=2;;i++)
	{
		f[i].len=f[i-1].len+f[i-2].len;
		for(int j=0;j<26;j++)
			f[i].a[j]=f[i-1].a[j]+f[i-2].a[j];
		if(f[i].len>qr)
		{
			maxn=i;
			break;
		}
	}
	data ans=query(maxn,qr)-query(maxn,ql-1);
	for(int i=0;i<26;i++)
		printf("%lld ",ans.a[i]);
	return 0;
}
/*
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
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
*/