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>
#define ll long long
#define fo(i,x,y) for(int i=x;i<=y;i++)
#define fd(i,x,y) for(int i=x;i>=y;i--)
using namespace std;

const int maxn=200007;

int n,Brd;
ll L,R;
char a[maxn];
int Fail[maxn];
ll Ans[26];
int AllCnt[26],BrdCnt[26];

void Init(){
	scanf("%s",a+1);
	scanf("%lld%lld",&L,&R);
	n=strlen(a+1)/2;
	int j=0;
	fo(i,2,n){
		while (j && a[j+1]!=a[i]) j=Fail[j];
		if (a[j+1]==a[i]) j++;
		Fail[i]=j;
	}
	Brd=n-Fail[n];
	fo(i,1,n) AllCnt[a[i]-'a']++;
	fo(i,1,Brd) BrdCnt[a[i]-'a']++;
}

void Calc(ll N,int Sig){
	if (N<=n){
		fo(i,1,N) Ans[a[i]-'a']+=Sig;
		return;
	}else fo(k,0,25) Ans[k]+=Sig*AllCnt[k];
	ll iSnum=1,iTnum=1,Res=N-n,iS=1,iT=0;
	while (N>iSnum*n+iTnum*Brd){
		Res=N-iSnum*n-iTnum*Brd;
		fo(k,0,25) Ans[k]+=Sig*(AllCnt[k]*(iSnum-iS)+BrdCnt[k]*(iTnum-iT));
		ll i=iS,j=iT;
		iS=iSnum,iT=iTnum;
		iSnum+=i;
		iTnum+=j;
	}
	if (Res)
		Calc(Res,Sig);
}
void Solve(){
	Calc(R,1);
	Calc(L-1,-1);
}

void Print(){
	fo(i,0,25) printf("%lld ",Ans[i]);
}

int main(){
	Init();
	Solve();
	Print();
	return 0;
}