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:
// luogu-judger-enable-o2
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define maxn 200010
using namespace std;
typedef long long ll;
char s[maxn];
ll L,R;
int n,T;
int fail[maxn];
void get_fail()
{
    fail[1]=0;
    for(int i=2;i<=n;i++)
    {
        int j=fail[i-1];
        while(j&&s[j+1]!=s[i])  j=fail[j];
        if(s[j+1]==s[i])  j++;
        fail[i]=j;
    }
    T=n-fail[n];
}
int sum[maxn][30];
ll calc(ll x,int c)
{
    if(x<=n)    return sum[x][c];
    if(x<=2*n)  return sum[n][c]+sum[x-n][c];
    ll f1=sum[n][c],f2=sum[n][c]+sum[T][c],l1=n,l2=n+T;
    while(x>l1+l2)
    {
        ll tmp=f2;
        f2+=f1;f1=tmp;
        tmp=l2;
        l2+=l1;l1=tmp;
    }
    return f2+calc(x-l2,c);
}
int main()
{
    scanf("%s",s+1);
    cin>>L>>R;
    n=strlen(s+1)/2;
    get_fail();
    for(int i=1;i<=n;i++)
    {
        memcpy(sum[i],sum[i-1],sizeof(sum[i-1]));
        sum[i][s[i]-'a']++;
    }
    for(int i=0;i<26;i++)  printf("%lld ",calc(R,i)-calc(L-1,i));
    puts("");
    return 0;
}