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>
#define ll long long
#define pi acos(-1)
#define inf 1e18
using namespace std;
const int N=200050;
inline ll rd(){
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	return f?x:-x;
}char str[N];int nxt[N],n,m,tot,top;
ll len[1005],nt[1005],cnt[1005][27],ct2[1005][27],o[27],ans[27];
inline void kmp(){
	int j=0;
	for(int i=2;i<=n;i++){
		while(j&&str[j+1]!=str[i]) j=nxt[j];
		if(str[j+1]==str[i]) j++;
		nxt[i]=j;
	}
}
inline void work(ll n,int f){
	for(int i=top;i>=1;i--){
		if(n>=len[i]){
			for(int j=0;j<26;j++) ans[j]+=f*(cnt[i][j]*2+ct2[i][j]);
			n-=len[i];
		}
	}
	if(n>=len[0]){
		for(int j=0;j<26;j++) ans[j]+=f*o[j];
		n-=len[0];
	}for(int i=1;i<=n;i++) ans[str[i]-'a']+=f;
}
inline void solve(ll n,int f){
	for(int i=0;i<26;i++) ans[i]+=1ll*f*(n/tot)*o[i];
	n%=tot;
	for(int i=1;i<=n;i++) ans[str[i]-'a']+=f; 
}
int main(){
	scanf("%s",str+1);n=strlen(str+1);n/=2;
	kmp();ll l=rd(),r=rd();
	if(n%(n-nxt[n])){
		for(int i=1;i<=nxt[n];i++) cnt[0][str[i]-'a']++;
		for(int i=nxt[n]+1;i<=n-nxt[n];i++) ct2[0][str[i]-'a']++; 
		for(int i=1;i<=n;i++) o[str[i]-'a']++;
		len[0]=n;nt[0]=nxt[n];
		if(nxt[n]>=(n+1)/2){
			top++;
			len[top]=len[top-1]*2-nt[top-1];
			nt[top]=len[top-1]-nt[top-1];
			for(int i=1;i<=nt[top];i++){
				cnt[top][str[i]-'a']++;
			}
			for(int i=n-nxt[n]+1;i<=n;i++){
				ct2[top][str[i]-'a']++; 
			} 
		} 
		while(len[top]<inf){
			top++;
			len[top]=len[top-1]*2-nt[top-1];
			nt[top]=len[top-1]-nt[top-1];
			for(int i=0;i<26;i++){
				cnt[top][i]=cnt[top-1][i]+ct2[top-1][i];
				ct2[top][i]=cnt[top-1][i];
			}
		}
		work(r,1);work(l-1,-1);
	}
	else{
		tot=n-nxt[n];
		for(int i=1;i<=tot;i++){
			o[str[i]-'a']++;
		}solve(r,1);solve(l-1,-1);
	}
	for(int i=0;i<26;i++) printf("%lld ",ans[i]);
	return 0;
}