Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define ll long long

using namespace std;

const int maxn=2e5+5;

ll a[maxn][28];
ll i,j,k,l,r,x,y,n,t;
char s[maxn];
int ne[maxn];
bool bz;

void ge_border(){
	fo(i,2,n){
		while (j && s[j+1]!=s[i]) j=ne[j];
		if (s[j+1]==s[i]) j++;
		ne[i]=j;
	}
	t=n-ne[n];
	if (n%t==0) bz=true;
}
ll calc(ll x,int c){
	if (x<=n) return a[x][c];
	if (x<=2*n) return a[n][c]+a[x-n][c];
	ll t1=n,t2=n+t,an1=a[n][c],an2=an1+a[t][c],t3,an3;
	while (t1+t2<=x){
		t3=t2; t2=t1+t2; t1=t3;
		an3=an2; an2=an1+an2; an1=an3;
	}
	return an2+calc(x-t2,c);
}
ll ge_ans(ll x,int c){
	if (x<=n) return a[x][c];
	if (bz){
		x=x-n; ll ans=a[n][c];
		ans=ans+(x/t)*a[t][c]+a[x%t][c];
		return ans; 
	}
	return calc(x,c);
}
int main(){
//	freopen("077f.in","r",stdin);
	scanf("%s",s+1);
	scanf("%lld%lld",&l,&r);
	n=strlen(s+1)/2;
	ge_border();
	fo(i,1,n){
		fo(j,1,26) a[i][j]=a[i-1][j];
		a[i][s[i]-96]++;
	}
	fo(j,1,26) printf("%lld ",-ge_ans(l-1,j)+ge_ans(r,j));
	return 0;
}