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<cstdio>
#include<cstring>
const int N=255555;
typedef long long LL;
struct data{
	LL len,a[26];
	inline void operator+=(const data&rhs){
		len+=rhs.len;
		for(int i=0;i<26;++i)a[i]+=rhs.a[i];
	}
	inline data operator-(const data&rhs)const{
		data c;
		c.len=0;
		for(int i=0;i<26;++i)c.a[i]=a[i]-rhs.a[i];
		return c;
	}
}d[500],ans;
char s[N];
int fail[N],n;
LL L,R;
void getfail(){
    for(int i=2,j=0;i<=n;++i){
        while(j&&s[j+1]!=s[i])j=fail[j];
        if(s[j+1]==s[i])++j;
        fail[i]=j;
    }
}
data solve(LL R){
	data ret;
	for(int i=ret.len=0;i<26;++i)ret.a[i]=0;
	if(n%(n-fail[n])==0){
		LL k=R/n;
		for(int i=1;i<=n;++i)ret.a[s[i]-'a']+=k;
		for(int i=1;i<=R%n;++i)++ret.a[s[i]-'a'];
	}else{
		memset(d,0,sizeof d);
		int tot=2;
		d[2].len=n;
		for(int i=1;i<=n;++i)++d[2].a[s[i]-'a'];
		d[1].len=n-fail[n];
		for(int i=1;i<=n-fail[n];++i)++d[1].a[s[i]-'a'];
		while(d[tot].len<=R){
			++tot;
			d[tot]=d[tot-1],d[tot]+=d[tot-2];
		}
		for(int i=tot;i;--i)
		if(R>=d[i].len)R-=d[i].len,ret+=d[i];
		for(int i=1;i<=R;++i)++ret.a[s[i]-'a'];
	}
	return ret;
}
int main(){
	scanf("%s%lld%lld",s+1,&L,&R);
	n=strlen(s+1)>>1;
	getfail();
	ans=solve(R);
	ans=ans-solve(L-1);
	for(int i=0;i<26;++i)
	printf("%lld ",ans.a[i]);
	return 0;
}