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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 200010
#define lint long long
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
char s[N];int nxt[N];
struct E{
	lint len,cnt[26];E() { len=0,memset(cnt,0,sizeof cnt); }
	inline E operator=(const E &e)
	{	return len=e.len,memcpy(cnt,e.cnt,sizeof cnt),*this;	}
	inline E operator+(const E &e)const
	{
		E res;res=*this;
		for(int i=0;i<26;i++) res.cnt[i]+=e.cnt[i];
		return res.len+=e.len,res;
	}
	inline E operator-(const E &e)const
	{
		E res;res=*this;
		for(int i=0;i<26;i++) res.cnt[i]-=e.cnt[i];
		return res.len-=e.len,res;
	}
	inline E operator+=(const E &e) { return (*this)=(*this)+e; }
}a[1010];
inline int getg(char *s,int n)
{
	for(int i=2;i<=n;i++)
	{
		int &j=nxt[i];j=nxt[i-1];
		while(j&&s[j+1]!=s[i]) j=nxt[j];
		if(s[j+1]==s[i]) j++;
	}
	return n-nxt[n];
}
inline E query(lint n,int c)
{
	E ans;
	for(int i=c;i>=1;i--)
		if(n>=a[i].len) n-=a[i].len,ans+=a[i];
	for(int i=1;i<=n;i++) ans.cnt[s[i]-'a']++;
	return ans;
}
int main()
{
	scanf("%s",s+1);int n=(int)strlen(s+1),c=2;
	int k=n/2,g=getg(s,k);lint l,r;cin>>l>>r;
	for(int i=1;i<=k;i++) a[1].cnt[s[i]-'a']++;
	a[2]=a[1],a[1].len=k,a[2].len=k+g;
	for(int i=1;i<=g;i++) a[2].cnt[s[i]-'a']++;
	while(a[c-1].len+a[c].len<=r) c++,a[c]=a[c-1]+a[c-2];
	E ans=query(r,c);
	ans=ans-query(l-1,c);
	for(int i=0;i<26;i++) printf("%lld ",ans.cnt[i]);
	return !printf("\n");
}