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<cstring>
#include<cstdio>
#define ll long long
#define MN 200000
#define mod 1000000007
using namespace std;
char s[MN+5];int n,cnt=2,ha[MN+5],pw[MN+5];
ll l,r,len[105],num[105][27],ans[27];
inline int Ha(int l,int r){return (ha[r]-1LL*ha[l-1]*pw[r-l+1]%mod+mod)%mod;}
void Calc(ll l,int x,int v)
{
	if(x<=1)
	{
		for(int j=1;j<=l;++j) ans[s[j]-'a']+=v;
		return;	
	}
	if(l<=len[x-1]) Calc(l,x-1,v);
	else 
	{
		for(int j=0;j<26;++j) ans[j]+=v*num[x-1][j];
		Calc(l-len[x-1],x-2,v);	
	}
}
int main()
{
	scanf("%s%lld%lld",s+1,&l,&r);n=strlen(s+1)/2;pw[0]=1;
	for(int i=1;i<=n;++i) pw[i]=31LL*pw[i-1]%mod,ha[i]=(31LL*ha[i-1]+s[i]-'a'+1)%mod;
	for(int i=2;i<=n+1;++i) if(i>n||Ha(i,n)==Ha(1,n-i+1))
	{
		len[1]=n;len[2]=n+i-1;
		for(int j=1;j<=n;++j) ++num[1][s[j]-'a'],num[2][s[j]-'a']+=1+(j<i);
		break;	
	}
	while(len[cnt]<r)
	{
		++cnt;len[cnt]=len[cnt-1]+len[cnt-2];
		for(int j=0;j<26;++j) num[cnt][j]=num[cnt-1][j]+num[cnt-2][j];
	}
	Calc(r,cnt,1);Calc(l-1,cnt,-1);
	for(int i=0;i<26;++i) printf("%lld ",ans[i]);
	return 0;
}

