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<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 201000
#define ll long long
using namespace std;
char s[N];
int nxt[N],n,m,t,a[N][26];
void kmp()
{
	int j=0;
	fo(i,2,m)
	{
		while(j&&s[j+1]!=s[i]) j=nxt[j];
		if(s[j+1]==s[i]) j++;
		nxt[i]=j;
	}
}
ll calc(int x,ll l)
{
	if(l<=m) return a[l][x];
	ll l1=t,l2=m,b1=a[t][x],b2=a[m][x];
	while(l2<=l)
	{
		ll jy=b1;
		b1=b2;b2=b2+jy;
		jy=l1;l1=l2;l2=l1+jy;
	}
	if(l==l1) return b1;
	else return b1+calc(x,l-l1);
}
int main()
{
	scanf("%s\n",s+1);
	n=strlen(s+1);
	m=n/2;
	kmp();
	t=m-nxt[m];
	fo(i,1,n)
	{
		fo(j,0,25) a[i][j]=a[i-1][j];
		a[i][s[i]-97]++;
	}
	ll l,r;scanf("%lld %lld",&l,&r);
	fo(i,0,25) printf("%lld ",calc(i,r)-calc(i,l-1));
}