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 <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 666666
int n,pl,g;
ll qz[SZ],hz[SZ];
char s[SZ]; ll l,r;
struct arr {ll c[26];}clr;
arr operator + (arr a,arr b)
{for(int i=0;i<26;++i) a.c[i]+=b.c[i]; return a;}
arr operator - (arr a,arr b)
{for(int i=0;i<26;++i) a.c[i]-=b.c[i]; return a;}
arr ps[SZ]; ll len[SZ];
arr calc(char*l,char*r)
{
	arr g=clr;
	while(l!=r) ++g.c[*(l++)-'a'];
	return g;
}
arr qry(int r,ll p)
{
	if(!p) return clr;
	if(r<=1) return calc(s+1,s+p+1);
	if(p<=len[r-1]) return qry(r-1,p);
	return ps[r-1]+qry(r-2,p-len[r-1]);
}
arr qry(ll p)
{return qry(g,p);}
int main()
{
	memset(&clr,0,sizeof clr);
	scanf("%s%lld%lld",s+1,&l,&r);
	n=strlen(s+1); n/=2; s[n+1]=0;
	for(int i=1;i<=n;++i)
		qz[i]=qz[i-1]*233+s[i];
	{
	ll po=1;
	for(int i=n;i>=1;--i)
		hz[i]=hz[i+1]+ll(s[i])*po,po*=233;
	}
	for(int i=1;i<=n;++i)
		if(qz[n-i]==hz[i+1])
		{pl=i; break;}
	len[0]=pl; len[1]=n;
	ps[0]=calc(s+1,s+pl+1);
	ps[1]=calc(s+1,s+n+1);
	for(int i=2;;++i)
	{
		len[i]=len[i-1]+len[i-2];
		ps[i]=ps[i-1]+ps[i-2];
		if(len[i]>1e18) {g=i; break;}
	}
	arr ans=qry(r)-qry(l-1);
	for(int i=0;i<26;++i)
		printf("%lld ",ans.c[i]);
}