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 N 200009
using namespace std;

int n,len,f[N]; ll l,r,ans[26]; char s[N];
int calc1(ll m,int p){
	int i,t=m%n; m/=n;
	for (i=1; i<=n; i++) ans[s[i]-'a']+=p*m;
	for (i=1; i<=t; i++) ans[s[i]-'a']+=p;
}
void calc2(ll m,int p){
	int i;
	if (m<=(n<<1)){
		for (i=1; i<=m; i++) ans[s[i]-'a']+=p;
		return;
	}
	ll x=n,y=f[n],A=1,B=1,a=1,b=0;
	for (; (x<<1)<=m; y=x-y,x+=y){
		a=A-a; b=B-b;
		A+=a; B+=b;
	}
	for (i=1; i<=f[n]; i++) ans[s[i]-'a']+=A*p;
	for (i=f[n]+1; i<=n; i++) ans[s[i]-'a']+=B*p;
	calc2(m-x,p);
}
int main(){
	scanf("%s",s+1); n=strlen(s+1)>>1;
	int i;
	for (i=2; i<=n; i++){
		f[i]=f[i-1];
		for (; f[i] && s[i]!=s[f[i]+1]; f[i]=f[f[i]]);
		if (s[i]==s[f[i]+1]) f[i]++;
	}
	scanf("%lld%lld",&l,&r);
	if (!(n%(n-f[n]))){
		n-=f[n];
		calc1(r,1); calc1(l-1,-1);
	} else{
		calc2(r,1); calc2(l-1,-1);
	}
	for (i=0; i<26; i++) printf("%lld%c",ans[i],(i<25)?' ':'\n');
	return 0;
}
