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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define MN 210000
using namespace std;
 
ll n,k,l,r,ne[MN],w[26];
char s[MN];
void work(ll r){
	if (r<=n) for (ll i=0;i<r;i++) w[s[i]-'a']++;else{
		ll _s=0,_t=1,S=1,T=0;
		for (;S*n+T*k<r;) S+=_s,T+=_t,_s=S-_s,_t=T-_t;
		work(r-_s*n-_t*k);
		for (ll i=0;i<k;i++) w[s[i]-'a']+=_t;
		for (ll i=0;i<n;i++) w[s[i]-'a']+=_s;
	}
}
int main(){
	scanf("%s%lld%lld",s,&l,&r);
	n=strlen(s)/2;ne[0]=-1;
	for (ll i=1;i<n;i++){
		ll j=ne[i-1];
		while (j!=-1&&s[i]!=s[j+1]) j=ne[j];
		ne[i]=j+(s[i]==s[j+1]);
	}
	k=n-ne[n-1]-1;
	work(l-1);
	for (ll i=0;i<26;i++) w[i]=-w[i];
	work(r);
	for (ll i=0;i<26;i++) printf("%lld ",w[i]);
}