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 <cstdlib>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,q) for(int i=A[q];i;i=B[i][0])
using namespace std;
typedef long long LL;
const int N=200500;
int read(int &n)
{
	char ch=' ';int q=0,w=1;
	for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());
	if(ch=='-')w=-1,ch=getchar();
	for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;
}
int m,n,ans;
int a[N],nx[N];
LL b[N][30],bv[N];
LL Ans[30];
void Doit(LL q,LL K)
{
	if(q<=n)
	{
		fo(i,1,q)Ans[a[i]]+=K;
		return;
	}
	fo(i,1,N)if(bv[i]>q)
	{
		fo(j,0,25)Ans[j]+=b[i-1][j]*K;
		Doit(q-bv[i-1],K);
		return;
	}
}
int main()
{
	int q;
	char ch=' ';
	for(ch=getchar();ch<='z'&&ch>='a';ch=getchar())a[++n]=ch-'a';
	n>>=1;
	q=0;
	fo(i,2,n)
	{
		for(;q&&a[q+1]!=a[i];q=nx[q]);
		if(a[q+1]==a[i])++q;
		nx[i]=q;
	}
	m=n-nx[n];
	fo(i,1,n)++b[0][a[i]];
	fo(i,0,25)b[1][i]=b[0][i];
	fo(i,1,m)++b[1][a[i]];
	bv[0]=n,bv[1]=n+m;
	fo(i,2,1000)
	{
		bv[i]=bv[i-1]+bv[i-2];
		fo(j,0,25)b[i][j]=b[i-1][j]+b[i-2][j];
	}
	LL l,r;
	scanf("%lld%lld",&l,&r);
	Doit(r,1);
	Doit(l-1,-1);
	fo(i,0,25)printf("%lld ",Ans[i]);
	return 0;
}