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<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)

using namespace std;
const int N=200005;
int n,m,c,s[N][30],f[N];
char S[N];
typedef long long LL;
LL l,r;

LL calc(LL x)
{
	if(x<=n) return s[x][c];
	LL s1=s[n][c],s0=s[m][c];
	LL l1=n,l0=m;
	while(l1<=x) swap(l1,l0),l1+=l0,swap(s1,s0),s1+=s0;
	return s0+calc(x-l0);
}

int main()
{
	scanf("%s",S+1),n=strlen(S+1)/2;
	rep(i,1,n)
		rep(j,0,25) s[i][j]=s[i-1][j]+(S[i]-'a'==j);
	rep(i,2,n)
	{
		int k=f[i-1];
		while(k && S[k+1]!=S[i]) k=f[k];
		if(S[k+1]==S[i]) ++k;
		f[i]=k;
	}
	m=n-f[n];
	scanf("%lld %lld",&l,&r);
	rep(i,0,25) c=i,printf("%lld ",calc(r)-calc(l-1));
	puts("");
	return 0;
}