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<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;

#define LL long long
#define DB double
#define MOD 1000000007
#define MAXN 1000000
#define INF 1000000
#define mem(x,v) memset(x,v,sizeof(x))

LL read(){
    LL x=0,F=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*F;
}

int n,m,nxt,res;LL l,r;
char s[MAXN+5],t[MAXN+5];
int f[MAXN+5],cnt[MAXN+5][30];

void getfail(int *fail){
    for(int i=2,j=0;i<=n;i++){
        while(j&&t[i]!=t[j+1])j=fail[j];
        j+=(t[i]==t[j+1]);
        fail[i]=j;
    }
}
LL getres(LL p,int c){
    if(p<=n)return cnt[p][c];
    LL res0=cnt[m][c],res1=cnt[n][c],L0=m,L1=n,tmp;
    while(L0+L1<p){
        tmp=L1,L1+=L0,L0=tmp;
        tmp=res1,res1+=res0,res0=tmp;
    }
    return res1+getres(p-L1,c);
}

int main(){
    scanf("%s",t+1);l=read(),r=read();
    n=strlen(t+1)/2;
    getfail(f),m=n-f[n];
    for(int c=0;c<26;c++)
        for(int i=1;i<=n;i++)
        cnt[i][c]=cnt[i-1][c]+(t[i]==c+'a');
    for(int x=0;x<26;x++)
    printf("%lld ",getres(r,x)-getres(l-1,x));
}