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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int p=998244353;
typedef unsigned long long ull;
int n,len,mx;
long long l,r,siz[110],cnt1[110],cnt2[110],ans[26],sum[110][26];
ull fhash[200010],bhash[200010],bs[200010];
char s[200010];
void solve(long long m,int type){
	for(int i=mx;~i;i--){
		if(m>=siz[i]){
			for(int j=0;j<26;j++)
				ans[j]+=type?sum[i][j]:-sum[i][j];
			m-=siz[i];
		}
	}
	for(int i=1;i<=m;i++)
		ans[s[i]-'a']+=type?1:-1;
	return;
}
int main(){
	scanf("%s%lld%lld",s+1,&l,&r);
	n=strlen(s+1);
	bs[0]=1;
	for(int i=1;i<=n;i++){
		fhash[i]=fhash[i-1]*p+(s[i]-'a'+1);
		bs[i]=bs[i-1]*p;
	}
	for(int i=n;~i;i--)
		bhash[i]=bhash[i+1]+(s[i]-'a'+1)*bs[n-i];
	for(int i=n/2-1;~i;i--)
		if(fhash[i]==bhash[n-i+1]){
			len=i;
			break;
		}
	for(int i=1;i<=n/2;i++)
		sum[0][s[i]-'a']++;
	for(int i=0;i<26;i++)
		sum[1][i]=sum[0][i];
	for(int i=1;i<=n/2-len;i++)
		sum[1][s[i]-'a']++;
	siz[0]=n/2;
	siz[1]=n-len;
	for(int i=2;i<=100;i++){
		for(int j=0;j<26;j++)
			sum[i][j]=sum[i-1][j]+sum[i-2][j];
		siz[i]=siz[i-1]+siz[i-2];
		if(siz[i]>1e18){
			mx=i;
			break;
		}
	}
	solve(l-1,0);
	solve(r,1);
	for(int i=0;i<26;i++)
		printf("%lld ",ans[i]);
	return 0;
}