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>
#define ll long long
const int N=200050;
using namespace std;
int gi(){
  char ch=getchar(); int x=0,q=0;
  while(ch<'0' || ch>'9') (ch=='-'?q=1:0),ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
char s[N];
int sum[N][26],n,nx[N],cir;
ll calc(ll m,int c){
  if (m<=n) return sum[m][c];
  if (n%cir==0)
    return 1ll*m/cir*sum[cir][c]+sum[m%cir][c];
  else{
    ll a=cir,b=n,sa=sum[cir][c],sb=sum[n][c];
    while(m>=a+b){
      b+=a,sb+=sa;
      a=b-a,sa=sb-sa;
    }
    return sb+calc(m-b,c);
  }
}
int main(){
  scanf("%s",s+1);
  n=strlen(s+1)/2;
  for (int i=1; i<=n; ++i){
    for (int j=0; j<26; ++j)
      sum[i][j]=sum[i-1][j];
    ++sum[i][s[i]-'a'];
  }
  for (int i=2; i<=n; ++i){
    int j=nx[i-1];
    while(j&&s[j+1]!=s[i]) j=nx[j];
    nx[i]=j+(s[j+1]==s[i]);
  }
  cir=n-nx[n];
  ll l,r;
  cin>>l>>r;
  for (int i=0; i<26; ++i)
    printf("%lld ",calc(r,i)-calc(l-1,i));
  return 0;
}