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<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int MX=400005;
const LL MXP=1000000000000000000ll;

char s[MX];int n;int bd[MX];
int T;
LL len[201],cf[201][26];
LL c[MX][26];
int typ;
#define A(x,y) for(int iA=0;iA<26;iA++)x[iA]+=y[iA]
void ini(){
	bd[0]=0;
	n=strlen(s);
	for(int i=1;i<n;i++){
		int t=bd[i-1];
		while(t&&s[t]!=s[i])t=bd[t-1];
		bd[i]=(s[t]==s[i]?t+1:0);
	}
	int l=bd[n-1];
	while(l>=n/2)l=bd[l-1];
	n-=l;
	memset(s+n,0,sizeof(char)*l);
	cerr<<s<<endl;
	T=n-bd[n-1];//最小正周期 
	cerr<<T<<endl;
	for(int i=0;i<n;i++){
		if(i)memcpy(c[i],c[i-1],sizeof(c[i]));
		c[i][s[i]-'a']++;
	}
	if(bd[n-1]%T){//fib
		typ=0;
		len[1]=n,len[2]=n+T;
		for(int i=0;i<26;i++)memcpy(cf[1],c[n-1],sizeof(cf[1]));
		memcpy(cf[2],cf[1],sizeof(cf[1]));
		A(cf[2],c[T-1]);
		for(int i=3;;i++){
			len[i]=len[i-1]+len[i-2];
			memcpy(cf[i],cf[i-1],sizeof(cf[i-1]));
			A(cf[i],cf[i-2]);
			if(len[i]>MXP)break;
		}
	}else{//cycle
		typ=1;
	}
}
LL cal(LL x,int ch){
	if(!x)return 0;
	if(typ)return (x%T?c[x%T-1][ch]:0)+c[T-1][ch]*(x/T);
	else{
		if(x<=n)return c[x-1][ch];
		for(int i=1;;i++){
			if(len[i+1]>x)return cf[i][ch]+cal(x-len[i],ch);
		}
	}
}
int main(){
	cin>>s;
	ini();
	LL L,R;cin>>L>>R;
	for(int i=0;i<26;i++)cout<<cal(R,i)-cal(L-1,i)<<' ';
	cout<<endl;
	return 0;
}