#include <cstring>
#include <cstdio>

typedef long long LL;
const int N=100005;

int cnt[N][26],fail[N],n,t;
char s[N*2];

LL calc(LL pos,int c){
	if (pos<=n) return cnt[pos][c];
	LL s1=cnt[n][c],s2=cnt[t][c],len1=n,len2=t;
	for (;len1+len2<=pos;){
		LL tmp=s1;s1+=s2;s2=tmp;
		tmp=len1;len1+=len2;len2=tmp;
	}
	return s1+calc(pos-len1,c);
}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	fail[0]=-1;
	for (int i=1,j=-1;i<=n;fail[i++]=++j)
		for (;j>=0 && s[j+1]!=s[i];j=fail[j]);
	LL l,r;
	scanf("%lld%lld",&l,&r);
	t=n-fail[n];
	for (int i=1;i<=n;i++){
		for (int j=0;j<26;j++) cnt[i][j]=cnt[i-1][j];
		cnt[i][s[i]-'a']++;
	}
	for (int i=0;i<26;i++)
		printf("%lld ",calc(r,i)-calc(l-1,i));
}
### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <cstring>
#include <cstdio>

typedef long long LL;
const int N=100005;

int cnt[N][26],fail[N],n,t;
char s[N*2];

LL calc(LL pos,int c){
	if (pos<=n) return cnt[pos][c];
	LL s1=cnt[n][c],s2=cnt[t][c],len1=n,len2=t;
	for (;len1+len2<=pos;){
		LL tmp=s1;s1+=s2;s2=tmp;
		tmp=len1;len1+=len2;len2=tmp;
	}
	return s1+calc(pos-len1,c);
}

int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	fail[0]=-1;
	for (int i=1,j=-1;i<=n;fail[i++]=++j)
		for (;j>=0 && s[j+1]!=s[i];j=fail[j]);
	LL l,r;
	scanf("%lld%lld",&l,&r);
	t=n-fail[n];
	for (int i=1;i<=n;i++){
		for (int j=0;j<26;j++) cnt[i][j]=cnt[i-1][j];
		cnt[i][s[i]-'a']++;
	}
	for (int i=0;i<26;i++)
		printf("%lld ",calc(r,i)-calc(l-1,i));
}
```