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;
const int N=2e5+10;
typedef long long ll;

char s[N];
ll l,r;
int sum[N][26],c,nxt[N],k,n;

ll cal(ll m) {
	if(m<=n) return sum[m][c];
	ll l1=n,l2=k,s1=sum[n][c],s2=sum[k][c];
	while(l1+l2<=m) {
		ll tmp=l1;l1+=l2;l2=tmp;
		tmp=s1;s1+=s2;s2=tmp;
	}
	return s1+cal(m-l1);
}

int main() {
	scanf("%s",s+1),n=strlen(s+1)/2;
	for(int i=1;i<=n;i++) {
		for(int j=0;j<26;j++) sum[i][j]=sum[i-1][j];
		++sum[i][s[i]-'a'];
	}
	for(int i=2,j=0;i<=n;i++) {
		while(j&&s[j+1]!=s[i]) j=nxt[j];
		if(s[j+1]==s[i]) ++j;
		nxt[i]=j;
	}
	k=n-nxt[n];cin>>l>>r;
	for(c=0;c<26;c++) cout<<cal(r)-cal(l-1)<<' ';
	return 0;
}
