Problem:
Score : 1100 points

Problem Statement

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).

Sample Input 1

abaaba
6 10

Sample Output 1

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

Since f(abaaba)=abaababaab, the first ten characters in f^{10^{100}}(S) is also abaababaab. Thus, the sixth through the tenth characters are abaab. In this string, a appears three times, b appears twice and no other letters appear, and thus the output should be 3 and 2 followed by twenty-four 0s.

Sample Input 2

xx
1 1000000000000000000

Sample Output 2

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

Sample Input 3

vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000

Sample Output 3

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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
#include<cstring>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int maxn=200000+10;
const ll inf=1000000000000000000;
char s[maxn];
ll len[1000],nt[1000],cnt1[1000][27],cnt2[1000][27],o[27],ans[27];
int nxt[maxn];
int i,j,k,t,n,m,tot,top;
ll l,r;
void kmp(){
	j=0;
	fo(i,2,n){
		while (j&&s[j+1]!=s[i]) j=nxt[j];
		if (s[j+1]==s[i]) j++;
		nxt[i]=j;
	}
}
void work(ll n,int f){
	int i,j;
	fd(i,top,0)
		if (n>=len[i]){
			fo(j,0,25) ans[j]+=f*(cnt1[i][j]*2+cnt2[i][j]);
			n-=len[i];
		}
	fo(i,1,n) ans[s[i]-'a']+=f;
}
void solve(ll n,int f){
	int i,j;
	fo(i,0,25) ans[i]+=(ll)f*(n/tot)*o[i];
	n%=tot;
	fo(i,1,n) ans[s[i]-'a']+=f;
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	n/=2;
	kmp();
	if (nxt[n]<(n+1)/2){
		fo(i,1,nxt[n]) cnt1[0][s[i]-'a']++;
		fo(i,nxt[n]+1,n-nxt[n]) cnt2[0][s[i]-'a']++;
		len[0]=n;nt[0]=nxt[n];
		while (len[top]<inf){
			top++;
			len[top]=len[top-1]*2-nt[top-1];
			nt[top]=len[top-1]-nt[top-1];
			fo(i,0,25){
				cnt1[top][i]=cnt1[top-1][i]+cnt2[top-1][i];
				cnt2[top][i]=cnt1[top-1][i];
			}
		}
		scanf("%lld%lld",&l,&r);
		work(r,1);
		work(l-1,-1);
		fo(i,0,25) printf("%lld ",ans[i]);
	}
	else{
		tot=n-nxt[n];
		fo(i,1,tot) o[s[i]-'a']++;
		scanf("%lld%lld",&l,&r);
		solve(r,1);
		solve(l-1,-1);
		fo(i,0,25) printf("%lld ",ans[i]);
	}
}