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
#include<bits/stdc++.h> 
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
const int maxn=2e5+5;
int i,j,c,ne[maxn];
char S[maxn];
ll num[maxn][26],l,r,T,n;
ll f(ll len){
	if (len<=n) return num[len][c];
	if (len<=n*2) return num[n][c]+num[len-n][c];
	ll f1=num[n][c],f2=num[n][c]+num[T][c],l1=n,l2=n+T;
	for(;len>l1+l2;)swap(f1,f2),f2+=f1,swap(l1,l2),l2+=l1;
    return f2+f(len-l2);
}
ll work(ll len){
	if (n%T==0) return len/T*num[T][c]+num[len%T][c];
    return f(len);
}
int main(){
	scanf("%s",S+1);
	n=strlen(S+1),n>>=1;
	fo(i,1,n) {
		fo(j,0,25) num[i][j]=num[i-1][j];
		num[i][S[i]-'a']++;
	}
	j=0;
	fo(i,2,n){
		for(;j&&S[j+1]!=S[i];j=ne[j]);
		if (S[j+1]==S[i]) j++;
		ne[i]=j;
	}
	T=n-j;
	scanf("%lld%lld",&l,&r);
	fo(c,0,25) printf("%lld ",work(r)-work(l-1));
}