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<bits/stdc++.h>
using namespace std;
#define int long long
int n;
int n2;
char s[200010];
int nxt[200010];
void kmp(){
	int j=0;
	for(int i=2;i<=n;++i){
		while(s[j+1]!=s[i]&&j)j=nxt[j];
		if(s[i]==s[j+1])j++;
		nxt[i]=j;
	}
}
int sum[200010][50];
int l,r;
int len[200010];
int up;
int ans[200010];
void solve(int x,int f){
    for(int i=up;i>=0;i--)
        if(x>=len[i]){
            for(int j=0;j<26;j++)
                ans[j]+=f*sum[i][j];
            x-=len[i];
        }
    for(int i=1;i<=x;i++)ans[s[i]-'a']+=f;
}
signed main(){
	scanf("%s%lld%lld",s+1,&l,&r);
	n=strlen(s+1);
	n>>=1;
	kmp();
	n2=nxt[n];
	//cout<<n2<<endl;
	for(int i=1;i<=n;++i){
		sum[0][s[i]-'a']++;
		sum[1][s[i]-'a']++;
	}
	len[0]=n;
	for(int i=1;i<=n-n2;++i){
		sum[1][s[i]-'a']++;
	}
	len[1]=2*n-n2;
	for(int i=2;i<=200000;++i){
		for(int j=0;j<26;++j){
			sum[i][j]=sum[i-1][j]+sum[i-2][j];
		}
		len[i]=len[i-1]+len[i-2];
		if(len[i]>r){
			up=i;
			break;
		}
	}
	solve(l-1,-1),solve(r,1);
	for(int i=0;i<26;++i){
		printf("%lld ",ans[i]);
	}
	puts(""); 
}