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:
/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;
const int N=2e5;

char str[N+5];
ll lef,rig;
ll fib[200],cnt[200][30],calA[30],calB[30];
int pi[N+5];
int n,m;

void Calc(ll *res,ll pos){
    int p=m;
    while(pos>fib[0]){
        while(fib[p]>pos) p--;
        for(int i=0;i<26;i++)
            res[i]+=cnt[p][i];
        pos-=fib[p];
    }
    for(int i=0;i<pos;i++)
        res[str[i]-'a']++;
}
void Excalc(ll *res,ll pos){
    for(int i=0;i<n;i++)
        res[str[i]-'a']+=pos/n;
    for(int i=pos%n-1;i>=0;i--)
        res[str[i]-'a']++;
}
int main(){
    scanf("%s%lld%lld",str,&lef,&rig);
    n=strlen(str);
    n>>=1;
    for(int i=1;i<n;i++){
        int j=pi[i-1];
        while(j && str[i]!=str[j])
            j=pi[j-1];
        if(str[i]==str[j]) j++;
        pi[i]=j;
    }
    if(n%(n-pi[n-1])==0){
        Excalc(calA,lef-1);
        Excalc(calB,rig);
    }
    else{
        fib[0]=n;fib[1]=n+n-pi[n-1];
        for(int i=0;i<n;i++){
            cnt[0][str[i]-'a']++;
            cnt[1][str[i]-'a']++;
            if(i<n-pi[n-1]) cnt[1][str[i]-'a']++;
        }
        for(int i=2;i<87;i++){
            fib[i]=fib[i-1]+fib[i-2];
            if(fib[i]>=rig) break;
            m=i;
            //printf("%d:%lld\n",i,fib[i]);
            for(int j=0;j<26;j++)
            	cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
        }
        Calc(calA,lef-1);Calc(calB,rig);
    }
    for(int i=0;i<26;i++)
        printf("%lld%c",calB[i]-calA[i],i==25? '\n':' ');
    return 0;
}