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
#define _GLIBCXX_IOSTREAM
#include<bits/stdc++.h>
#define maxn 200100
using namespace std;
typedef long long ll;
char s[maxn];
int fail[maxn],n;
ll L,R,c[50],d[100][50],len[100];
void solve(ll x,int k,int flag){
	if(n%k==0){
		for(int i=1;i<=n&&i<=x;++i)
			c[s[i]-'a']+=flag;
		ll y=(x-n)/k;
		if(y>0){
			for(int i=1;i<=k;++i)c[s[i]-'a']+=flag*y;
			for(int i=1;i<=x-n-k*y;i++)c[s[i]-'a']+=flag;
		}
	} else {
		memset(d,0,sizeof(d)),memset(len,0,sizeof(len));
		for(int i=1;i<=k;++i)d[1][s[i]-'a']+=flag;len[1]=k;
		for(int i=1;i<=n;++i)d[2][s[i]-'a']+=flag;len[2]=n;
		for(int i=3;len[i-1]+len[i-2]<=x;i++){
			for(int j=0;j<26;++j)d[i][j]=d[i-1][j]+d[i-2][j];
			len[i]=len[i-1]+len[i-2];
//			printf("[Len:%lld]\n",len[i]);
		}
		int i=1;
		for(;len[i]<=x&&len[i];i++);i--;
		for(int j=i;j>=2;--j)if(x>=len[j]){
			x-=len[j];
			for(int k=0;k<26;++k)c[k]+=d[j][k];
		}
//		printf("[%lld]",x);
		for(int i=1;i<=x;++i)c[s[i]-'a']+=flag;
	}
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	fail[1]=0;
	for(int i=2;i<=n;++i){
		int j=fail[i-1];
		while(j&&s[j+1]!=s[i])j=fail[j];
		if(s[j+1]==s[i])fail[i]=j+1;
		else fail[i]=j;
	}
	scanf("%lld%lld",&L,&R);
	solve(L-1,n-fail[n],-1);
	solve(R,n-fail[n],1);
	for(int i=0;i<26;++i)printf("%lld ",c[i]);
}