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 <bits/stdc++.h>
using namespace std;
const int N=210000;
#define ll long long
char s[N];
ll L,R;
int nex[N];
ll len[210];
struct node
{
	ll a[26];
	void init(int x)
	{
		for(int i=1;i<=x;i++)
			a[s[i]-'a']++;
	}
	friend node operator + (const node &r1,const node &r2)
	{
		node ret;
		for(int i=0;i<26;i++)
			ret.a[i]=r1.a[i]+r2.a[i];
		return ret;
	}
	friend node operator - (const node &r1,const node &r2)
	{
		node ret;
		for(int i=0;i<26;i++)
			ret.a[i]=r1.a[i]-r2.a[i];
		return ret;
	}
	void print()
	{
		for(int i=0;i<26;i++)
			printf("%lld ",a[i]);
		puts("");
	}
}v[210];
node calc(ll x)
{
	int p;
	for(p=1;;p++)
		if(len[p]>=x)break;
	node ret,t;
	memset(&ret,0,sizeof(ret));
	memset(&t,0,sizeof(t));
	for(;p>=1;p--)
		if(len[p]<=x)
		{
			ret=ret+v[p];
			x-=len[p];
		}
	t.init(x);
	ret=ret+t;
	return ret;
}
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%s%lld%lld",s+1,&L,&R);
	len[2]=strlen(s+1)/2;
	nex[1]=0;
	for(int i=2,j=0;i<=len[2];i++)
	{
		while(j&&s[j+1]!=s[i])
			j=nex[j];
		if(s[j+1]==s[i])j++;
		nex[i]=j;
	}
	len[1]=len[2]-nex[len[2]];
	v[1].init(len[1]);
	v[2].init(len[2]);
	for(int i=3;;i++)
	{
		len[i]=len[i-1]+len[i-2];
		v[i]=v[i-1]+v[i-2];
		if(len[i]>=R)break;
	}
	(calc(R)-calc(L-1)).print();
	return 0;
}
