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 <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int ALP='z'-'a'+1,N=100;
string S;
ll l,r;
string gs;
ll len[N];
ll sch[N][ALP];
ll ans[ALP];
ll dfs(int n,int c,ll ub){
	ll ret=0;
	if (n==0){
		REP(i,min(ub,(ll)S.size())){
			if (S[i]==c+'a'){
				ret++;
			}
		}
	}else if (n==1){
		REP(i,min(ub,(ll)gs.size())){
			if (gs[i]==c+'a'){
				ret++;
			}
		}
	}else{
		if (ub<=len[n-1]){
			ret=dfs(n-1,c,ub);
		}else{
			ret+=sch[n-1][c]+dfs(n-2,c,ub-len[n-1]);
		}
	}
	return ret;
}
int main(){
	cin>>S>>l>>r;
	S.resize((int)S.size()/2);
	{
		vector<ll> kmp(S.size()+1);
		kmp[0]=-1;
		int j=-1;
		REP(i,(int)S.size()){
			while(j>=0 && S[j]!=S[i]) j=kmp[j];
			j++;
			kmp[i+1]=j;
		}
		int per=(int)S.size()-kmp[(int)S.size()];
		gs=S+S.substr(0,per);
	}
	len[0]=(int)S.size();
	len[1]=(int)gs.size();
	REP(j,len[0]){
		sch[0][S[j]-'a']++;
	}
	REP(i,len[1]){
		sch[1][gs[i]-'a']++;
	}
	{
		int i=1;
		do{
			i++;
			len[i]=len[i-1]+len[i-2];
			REP(c,ALP){
				sch[i][c]=sch[i-1][c]+sch[i-2][c];
			}
		}while(len[i]<r);
		REP(c,ALP){
			ans[c]=dfs(i,c,r);
		}
		REP(c,ALP){
			ans[c]-=dfs(i,c,l-1);
		}
	}
	REP(c,ALP){
		if (c){
			cout<<" ";
		}
		cout<<ans[c];
	}
	cout<<endl;
	return 0;
}