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<iostream>
#include<algorithm>
using namespace std;
#define next DCXISSOHANDSOME
#define ll long long

ll num[100010][26],next[100010],n,l,r,hh;
char s[200010];

inline ll rd()
{
	ll x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline ll fib(ll len,ll x)
{
	if (len<=n) return num[len][x];
	if (len<=n+hh) return num[n][x]+num[len-n][x];
	ll f1=num[n][x],f2=num[n][x]+num[hh][x],l1=n,l2=n+hh;
	while (len>=l1+l2)
	{
		ll hhh=f2;f2+=f1;f1=hhh;
		hhh=l2;l2+=l1;l1=hhh;
	}
	return f2+fib(len-l2,x);
}

inline ll calc(ll len,ll x)
{
	if (len<=n) return num[len][x];
	return fib(len,x);
}

int main()
{
	scanf("%s",s+1);
	n=strlen(s+1)>>1;
	l=rd();r=rd();
	next[1]=0;
	for (ll i=2;i<=n;i++)
	{
		ll j=next[i-1];
		for (;j&&s[j+1]!=s[i];j=next[j]);
		if (s[j+1]==s[i]) j++;
		next[i]=j;
	}
	for (ll i=1;i<=n;i++)
	{
		for (ll j=0;j<26;j++) num[i][j]=num[i-1][j];
		num[i][s[i]-'a']++;
	}
	hh=n-next[n];
	for (ll i=0;i<26;i++) printf("%lld ",calc(r,i)-calc(l-1,i));
	puts("");
	return 0;
}