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<bits/stdc++.h>
using namespace std;

const int N=400100,M=10000;
typedef long long ll;
int f[N],n;
ll tmp[M][36],len[M],ans[36],l,r;
char s[N];

void MP(){
	int j=f[0]=-1;f[1]=0;
	for(int i=1;i<n;f[++i]=++j)
		for(;~j&&s[i]!=s[j];j=f[j]);
}

void solve1(ll lim,int op){
	memset(tmp,0,sizeof tmp);
	for(int j=0;j<n+n-f[n];++j){
		if(j<n)tmp[0][s[j]]++;
		tmp[1][s[j]]++;
	}
	len[0]=n;len[1]=n+n-f[n];
	int i=2;
	for(;len[i-1]<=lim;++i){
		len[i]=len[i-1]+len[i-2];
		for(int j=0;j<26;++j)tmp[i][j]=tmp[i-1][j]+tmp[i-2][j];
	}
	for(i-=2;i>=0;--i)
		if(lim>=len[i]){
			for(int k=0;k<26;++k)ans[k]+=tmp[i][k]*op;
			lim-=len[i];
		}
	for(int j=0;j<lim;++j)ans[s[j]]+=op;
}

void solve2(ll lim,int op){
	--lim;
	for(int i=0;i<(n-f[n]);++i)
		ans[s[i]]+=(lim+n-f[n]-i)/(n-f[n])*op;
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>s>>l>>r;n=strlen(s)/2;
	for(int i=0;i<2*n;++i)s[i]-='a';MP();
	if(n%(n-f[n])){solve1(r,1);solve1(l-1,-1);}
	else{solve2(r,1);solve2(l-1,-1);}
	for(int i=0;i<26;++i)cout<<ans[i]<<" \n"[i==25];
	return 0;
}
