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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const ll MAX=1000040,MAX2=200;
char s[MAX],s1[MAX],s2[MAX];
ll li[MAX],li2[MAX],len,len1,len2,cnt[MAX2][26],fib[MAX2],ans[MAX],l,r;
void build_cnt(ll x,ll lim,ll w){
    for(int i=0;i<lim;i++){
        if(x<=fib[i]){
            if(i==1){
                for(int j=0;j<x;j++) ans[s2[j]-'a']+=w;
                return;
            }
            if(i==0){
                for(int j=0;j<x;j++) ans[s1[j]-'a']+=w;
                return;
            }
            for(int j=0;j<26;j++) ans[j]+=cnt[i-1][j]*w;
            build_cnt(x-fib[i-1],i-1,w);
            return;
        }
    }
}
int main(){
    scanf(" %s %lld %lld",s,&l,&r);

    len=strlen(s);
    for(ll i=1,j=0;i<len;i++){
        while((s[i]!=s[j]&&j!=0)) j=li[j];
        if(s[i]==s[j]) li[i+1]=++j;
    }
    for(ll i=2,j=0;i<len;i++){
        while((s[i]!=s[j]&&j!=0)||i+1<=2*(j+1)) j=li[j];
        if(s[i]==s[j]) li2[i+1]=++j;
    }
    for(ll i=0;i<len-li2[len]-li[len-li2[len]];i++) s2[i]=s[i];
    for(ll i=0;i<li[len-li2[len]];i++) s1[i]=s[i+len-li2[len]-li[len-li2[len]]];
    len1=strlen(s1),len2=strlen(s2);
    fib[0]=len1,fib[1]=len2;
    for(ll i=0;i<len1;i++) cnt[0][s1[i]-'a']++;
    for(ll i=0;i<len2;i++) cnt[1][s2[i]-'a']++;
    for(ll i=2;i<MAX2;i++) {
        fib[i]=fib[i-1]+fib[i-2];
        for(ll j=0;j<26;j++) cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
    }
    build_cnt(r,MAX2,1);
    build_cnt(l-1,MAX2,-1);
    //printf("%d\n",strlen(s)/2);
    //printf("%s %s\n",s1,s2);
    for(ll i=0;i<26;i++) printf("%lld ",ans[i]);
}