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>

using namespace std;

typedef long long LL;

const int N=200005;
const int M=100;
const int C=26;

int app[N][C];
LL cnt[M][C];
int nxt[N];
LL len[M];
char s[N];
LL l,r;
int n,tot;

void pre()
{
	len[0]=n-nxt[n-1],len[1]=n;
	for (int i=2;len[i-1]<=r;++i)
	{
		len[tot=i]=len[i-1]+len[i-2];
		for (int c=0;c<C;++c) cnt[i][c]=cnt[i-1][c]+cnt[i-2][c];
	}
}

LL calc(LL m,int c,int cur)
{
	if (m<=n) return app[m][c];
	for (;len[cur]>m;--cur);
	return cnt[cur][c]+calc(m-len[cur],c,cur-1);
}

int main()
{
	//freopen("ss.in","r",stdin),freopen("ss.out","w",stdout);
	scanf("%s",s),n=strlen(s)>>1,scanf("%lld%lld",&l,&r);
	for (int i=1,ptr=0;i<n;++i)
	{
		for (;ptr&&s[i]!=s[ptr];ptr=nxt[ptr-1]);
		nxt[i]=ptr+=s[i]==s[ptr];
	}
	for (int i=1;i<=n;++i)
	{
		for (int c=0;c<C;++c) app[i][c]=app[i-1][c];
		++app[i][s[i-1]-'a'];
	}
	for (int c=0;c<C;++c) cnt[0][c]=app[n-nxt[n-1]][c],cnt[1][c]=app[n][c];
	pre();
	for (int c=0;c<C;++c) printf("%lld%c",calc(r,c,tot)-calc(l-1,c,tot),"\n "[c+1<C]);
	//fclose(stdin),fclose(stdout);
	return 0;
}