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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const double EPS=1e-10;
const double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
char in[2010000];
int fail[2010000];
int n;
map<pair<long long,long long>,long long>M;
long long calc(long long a,long long b,long long c,int d){
	if(M.count(make_pair(a,b)))return M[make_pair(a,b)];
	//printf("%lld %lld %lld %d\n",a,b,c,d);
	long long ret=0;
	if(n>=b){
		for(int i=0;i<a;i++){
			if('a'+d==in[i])ret++;
		}
		return M[make_pair(a,b)]=ret;
	}
	long long to=b-c;
	long long to2=c-to;
	ret+=calc(min(a,c),c,to,d);
	if(a>c){
		ret+=calc(a-c,to,to2,d);
	}
	return M[make_pair(a,b)]=ret;
}
int main(){
	scanf("%s",in);
 
	n=strlen(in)/2;
 
	for(int i=1;i<n;i++){
		int at=fail[i];
		while(1){
			if(in[at]==in[i]){
				fail[i+1]=at+1;
				break;
			}
			if(at==0)break;
			at=fail[at];
		}
	}
	long long L,R;
	scanf("%lld%lld",&L,&R);
	L--;
	int sh=n-fail[n];
	long long f1=n;
	long long f2=sh;
	while(f1<R){
		long long tmp=f1+f2;
		f2=f1;f1=tmp;
	}
	for(int i=0;i<26;i++){
		M.clear();
		printf("%lld ",calc(R,f1,f2,i)-calc(L,f1,f2,i));
	}
	printf("\n");
}