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 <bits/stdc++.h>
using namespace std;

typedef long long  LL;
int n,nxt[200010],k;
LL l,r,ans[26];
char s[200010];

void calc(LL m)
{
	if (m<=n)
		{
			for (int i=1; i<=m; i++)  ans[s[i]-'a']++;
			return;
		}
	LL a0=1,b0=0,c0=0,a1=1,b1=1,c1=1,a2,b2,c2;
	while (a1*k+b1*(n-k)<=m)
		{
			a2=a1+a0,b2=b1+b0,c2=c0;
			a0=a1,b0=b1,c0=c1,a1=a2,b1=b2,c1=c2;
		}
	for (int i=1; i<=k; i++)  ans[s[i]-'a']+=a0;
	for (int i=k+1; i<=n; i++)  ans[s[i]-'a']+=b0;
	calc(m-a0*k-b0*(n-k));
}

void work()
{
	scanf("%s\n%lld %lld",s+1,&l,&r),n=strlen(s+1)/2,l--;
	nxt[1]=0;
	for (int i=2,j=0; i<=n; i++)
		{
			while ((j)&&(s[i]!=s[j+1]))  j=nxt[j];
			if (s[i]==s[j+1])  nxt[i]=++j;  else  nxt[i]=j;
		}
	k=n-nxt[n];
	calc(l);
	for (int i=0; i<26; i++)  ans[i]*=-1;
	calc(r);
	for (int i=0; i<26; i++)  printf("%lld ",ans[i]);
}

int main()
{
	work();
	return 0;
}
