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 N 1111116
#define next nxt

int n,k,next[N];
char s[N];
long long l,r,ans[256];

void solve(long long m,long long t){
	if (m<=n){
		for (int i=1;i<=m;i++) ans[s[i]]+=t;
		return;
	}
	long long a0=1,b0=0,a1=1,b1=1;
	while (a1*k+b1*(n-k)<=m){
		a1+=a0;a0=a1-a0;
		b1+=b0;b0=b1-b0;
	}
	for (int i=1;i<=k;i++) ans[s[i]]+=t*a0;
	for (int i=k+1;i<=n;i++) ans[s[i]]+=t*b0;
	solve(m-a0*k-b0*(n-k),t);
}

long long read(){
	long long x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	scanf(" %s",s+1);n=strlen(s+1)/2;
	l=read();r=read();
	for (int i=2,j=0;i<=n;i++){
		while (j&&s[i]!=s[j+1]) j=next[j];
		next[i]=(j+=(s[i]==s[j+1]));
	}
	k=n-next[n];solve(l-1,-1);solve(r,1);
	for (int i='a';i<='z';i++)
		printf((i!='z')?"%lld ":"%lld\n",ans[i]);
	return 0;
}