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 <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define LL long long
#define N 200005
using namespace std;
int nx[N],n;
LL a[200][26],le[200],l,r;
char ch[N];
void kmp()
{
    int j=0;
    fo(i,2,n/2)
    {
        while(j&&ch[j+1]!=ch[i]) j=nx[j];
        if(ch[j+1]==ch[i]) j++;
        nx[i]=j;
    }
}
LL calc(int c,LL w)
{
    int id=1;
    if(le[id]>=w) 
    {
        int s=0;
        fo(i,1,w)
        {
            if(ch[i]-'a'==c) s++;
        }
        return s;
    }
    while(w>=le[id])
    {
        a[id+1][c]=a[id][c]+a[id-1][c];
        le[id+1]=le[id]+le[id-1];
        id++;
    }
    LL s=a[id-1][c];
    if(w==le[id-1]) return s;
    else return s+calc(c,w-le[id-1]);
}
int main()
{
    scanf("%s",ch+1);
    n=strlen(ch+1);
    kmp();
    cin>>l>>r;
    le[0]=n/2-nx[n/2];
    le[1]=n/2;
    fo(i,1,n/2) 
    {
        a[1][ch[i]-'a']++; 
        if(i<=le[0]) a[0][ch[i]-'a']++;
    }
    fo(i,0,25) printf("%lld ",calc(i,r)-calc(i,l-1));
}