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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=4e5+100;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
char s[N];
int nxt[N];
LL ans[26],cnt[1000][26],len[1000];
//每次会把串SS变成STST，其中T为S的最小周期
//如果T是S的约数，那么最后的串为STTTTTTTT的形式，可以直接计算
//如果T不是S的约数，那么串会变成 S T ST STS 这样的斐波那契形式
inline void solve(int n,int v,LL m) {
	int i,t;
	if (n%v) {
		memset(cnt,0,416);
		len[0]=v;
		for (i=1;i<=v;i++) cnt[0][s[i]-'a']++;
		len[1]=n;
		for (i=1;i<=n;i++) cnt[1][s[i]-'a']++;
		for (i=2;(len[i]=len[i-1]+len[i-2])<=m;i++)
			for (t=0;t<26;t++) cnt[i][t]=cnt[i-1][t]+cnt[i-2][t];
		while (i--)
			if (m>=len[i])
				for (t=0,m-=len[i];t<26;t++) ans[t]+=cnt[i][t];
		for (i=1;i<=m;i++) ans[s[i]-'a']++;
	} else {
		for (i=1;i<=n&&i<=m;i++)
			ans[s[i]-'a']++;
		if (m<=n) return;
		m-=n;
		for (i=1;i<=v;i++)
			ans[s[i]-'a']+=m/v+(m%v>=i);
	}
}
int main()
{
	scanf("%s",s+1);
	int n=strlen(s+1)/2,i,k,v;LL l,r;cin>>l>>r;
	for (i=2,k=0;i<=n;i++) {
		while (k&&s[k+1]!=s[i]) k=nxt[k];
		nxt[i]=k+=s[k+1]==s[i];
	}
	v=n-nxt[n];
	solve(n,v,l-1);
	for (i=0;i<26;i++) ans[i]*=-1;
	solve(n,v,r);
	for (i=0;i<26;i++)
		cout<<ans[i]<<" ";
	return 0;
}
