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<algorithm>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=22e4;

char s[N];
int len,n,m;
ll L,R;
int app[N][26],p[N];

ll count(ll a,ll b)
{
	if(a<=(n<<1))return a<=n?app[a][b]:app[n][b]+app[a-n][b];
	ll f1=n,f2=m,k1=app[n][b],k2=app[m][b],k3,f3;
	while(f1*2<a){
		f3=f1+f2; f2=f1; f1=f3;
		k3=k1+k2; k2=k1; k1=k3;
	}
	return k3+count(a-f1,b);
}

int main()
{
	scanf("%s",s+1);
	len=strlen(s+1); n=len>>1;
	cin>>L>>R; --L;
	fo(i,1,n){
		fo(l,0,25)app[i][l]=app[i-1][l];
		++app[i][s[i]-97];
	}
	for(int j=0,i=2;i<=n;++i){
		while(j!=0&&s[j+1]!=s[i])j=p[j];
		if(s[j+1]==s[i])++j;
		p[i]=j;
	}
	m=n-p[n];
	fo(i,0,25)printf("%lld ",count(R,i)-count(L,i));
}