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;
typedef long long ll;
const int N=2e5+10;
char s[N];
ll l,r,len,border,sta[N],suf[N],ans[N];
int n,top,nxt[N];
void query(ll len,int d){
	map<ll,ll> m;
	m[-len]=1;
	for (int i=top;i;i--){
		ll T=sta[i];
		for (map<ll,ll>::iterator i=m.begin();i!=m.end();++i){
			ll len=-i->first,cnt=i->second;
			if (len<=T) continue;
			m[-T]+=cnt*(len/T);
			m[-len%T]+=cnt;
		}
		while (-m.begin()->first>T) m.erase(m.begin());
	}
	for (map<ll,ll>::iterator i=m.begin();i!=m.end();++i) suf[-i->first]+=d*i->second;
}
int main()
{
	scanf("%s%lld%lld",s+1,&l,&r);
	n=len=sta[top=1]=strlen(s+1);
	for (int i=2;i<=n;i++){
		int j=nxt[i-1];
		while (j&&s[j+1]!=s[i]) j=nxt[j];
		nxt[i]=(s[j+1]==s[i]?j+1:0);
	}
	border=nxt[n/2];
	while (len<r){
		while (sta[top]>len-border) top--;
		sta[++top]=len-border;
		sta[++top]=2*(len-border);
		border=len/2-border;
		len=sta[top];
	}
	query(r,1);query(l-1,-1);
	for (int i=n;i;i--) suf[i]+=suf[i+1],ans[s[i]]+=suf[i];
	for (int i='a';i<='z';i++) printf("%lld ",ans[i]);
	puts("");
	return 0;
}