Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int n,m,fail[200010];
ll ans[30],len[95],sum[95][30],cnt[200010][30];
char s[200010];
void kmp()
{
	fail[0]=-1;
	int p=-1;
	for (int i=1;i<n;i++)
	{
		while (p>=0 && s[i]!=s[p+1])
		{
			p=fail[p];
		}
		if (s[i]==s[p+1])
		{
			p++;
		}
		fail[i]=p;
	}
}
void solve(ll k,ll f)
{
	ll p=k/(ll)m;
	for (int i=0;i<26;i++)
	{
		ans[i]+=f*p*cnt[m][i];
	}
	ll q=k%(ll)m;
	for (int i=0;i<26;i++)
	{
		ans[i]+=f*cnt[q][i];
	}
}
void work(ll k,ll f)
{
	int i=1;
	while (len[i]<=k)
	{
		i++;
	}
	for (;i;i--)
	{
		if (k>len[i])
		{
			for (int j=0;j<26;j++)
			{
				ans[j]+=f*sum[i][j];
			}
			k-=len[i];
		}
	}
	for (i=0;i<26;i++)
	{
		ans[i]+=f*cnt[k][i];
	}
}
int main()
{
	scanf("%s",&s);
	n=strlen(s);
	n/=2;
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<26;j++)
		{
			cnt[i][j]=cnt[i-1][j];
		}
		cnt[i][s[i-1]-'a']++;
	}
	kmp();
	m=n-1-fail[n-1];
	ll l,r;
	cin>>l>>r;
	if (n%m==0)
	{
		solve(r,1);
		solve(l-1,-1);
	}
	else
	{
		len[1]=n;
		len[2]=m+n;
		for (int i=0;i<26;i++)
		{
			sum[1][i]=cnt[n][i];
			sum[2][i]=cnt[m][i]+sum[1][i];
		}
		for (int i=3;i<=90;i++)
		{
			len[i]=len[i-2]+len[i-1];
			for (int j=0;j<26;j++)
			{
				sum[i][j]=sum[i-1][j]+sum[i-2][j];
			}
		}
		work(r,1);
		work(l-1,-1);
	}
	for (int i=0;i<26;i++)
	{
		printf("%lld ",ans[i]);
	}
	return 0;
}