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;
typedef long long ll;
ll n,sz[105],cnt[105][26],ans[26];
int nxt[200005],a[200005];
char s[200005];
void query(ll now,ll c)
{
	//printf("query(%lld,%lld)\n",now,c);
	if(now<=n)
	{
		for(int i=1;i<=now;i++) ans[a[i]]+=c;
		return;
	}
	int i=1;
	while(sz[i]<=now&&sz[i]>0) i++;
	for(int j=0;j<26;j++) ans[j]+=cnt[i-1][j]*c;
	query(now-sz[i-1],c);
}
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1;i<=n;i++) a[i]=s[i]-'a';
	n/=2;
	for(int i=2,j=0;i<=n;i++)
	{
		while(j&&a[j+1]!=a[i]) j=nxt[j];
		if(a[i]==a[j+1]) j++;
		nxt[i]=j;
		//printf("nxt[%d]=%d\n",i,j);
	}
	sz[0]=n,sz[1]=n+n-nxt[n];
	//printf("%lld %lld\n",sz[0],sz[1]);
	for(int i=1;i<=n;i++) cnt[0][a[i]]++,cnt[1][a[i]]++;
	for(int i=1;i<=n-nxt[n];i++) cnt[1][a[i]]++;
	for(int i=2;i<=100;i++)
	{
		sz[i]=sz[i-1]+sz[i-2];
		//printf("sz[%d]=%lld\n",i,sz[i]);
		for(int j=0;j<26;j++) cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
	}
	ll l,r;
	scanf("%lld%lld",&l,&r);
	query(r,1);
	query(l-1,-1);
	for(int i=0;i<26;i++) printf("%lld ",ans[i]);
	return 0; 
}