#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=200005;
int n,n2,ne[N],g[N][26];
char s[N];
ll l,r;
ll cal(int now,ll m){
	if (m<=n)return g[m][now];
	ll l1=n,l2=n2,k,s1=g[l1][now],s2=g[l2][now];
	while (l1+l2<=m){
		k=l1;l1+=l2;l2=k;
		k=s1;s1+=s2;s2=k;
	}
	return s1+cal(now,m-l1);
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	ne[0]=-1;
	for (int i=1,j=-1;i<=n;i++){
		while (j>=0&&s[j+1]!=s[i])j=ne[j];
		j++;ne[i]=j;
	}
	n2=n-ne[n];
	for (int i=1;i<=n;i++){
		for (int j=0;j<26;j++)g[i][j]=g[i-1][j];
		g[i][s[i]-97]++;
	}
	scanf("%lld%lld",&l,&r);
	for (int now=0;now<26;now++)printf("%lld ",cal(now,r)-cal(now,l-1));
}

### Prompt

Please formulate a Cpp solution to 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=200005;
int n,n2,ne[N],g[N][26];
char s[N];
ll l,r;
ll cal(int now,ll m){
	if (m<=n)return g[m][now];
	ll l1=n,l2=n2,k,s1=g[l1][now],s2=g[l2][now];
	while (l1+l2<=m){
		k=l1;l1+=l2;l2=k;
		k=s1;s1+=s2;s2=k;
	}
	return s1+cal(now,m-l1);
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	ne[0]=-1;
	for (int i=1,j=-1;i<=n;i++){
		while (j>=0&&s[j+1]!=s[i])j=ne[j];
		j++;ne[i]=j;
	}
	n2=n-ne[n];
	for (int i=1;i<=n;i++){
		for (int j=0;j<26;j++)g[i][j]=g[i-1][j];
		g[i][s[i]-97]++;
	}
	scanf("%lld%lld",&l,&r);
	for (int now=0;now<26;now++)printf("%lld ",cal(now,r)-cal(now,l-1));
}

```