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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define next Next
#define last Last
#define int long long
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
	int ret=0,f=0;char c=gc();
	while(!isdigit(c)){if(c=='-')f=1;c=gc();}
	while(isdigit(c)){ret=ret*10+c-48;c=gc();}
	if(f)return -ret;return ret;
}
const int N=200005;
int n,d,next[N];
int l,r,s[N][30];
char a[N];
void KMP()
{
	int k=0;
	for(int i=2;i<=n;i++)
	{
		for(;k&&a[k+1]!=a[i];)k=next[k];
		next[i]=(k+=(a[k+1]==a[i]));
	}
	d=n-next[n];
}
int fib(int p,int c)
{
	if(p<=n)return s[p][c];
	if(p<=n*2)return s[n][c]+s[p-n][c];
	int ans1=s[n][c]+s[d][c],ans2=s[n][c];
	int a=n+d,b=n;
	for(;p>a+b;)
	{
		int c=ans1;
		ans1+=ans2; 
		ans2=c;
		c=a;
		a+=b;
		b=c;
	}
	return ans1+fib(p-a,c);
}
int suan(int p,int c)
{
	if (p<=n) return s[p][c];
	if(n%d==0){
		int ans=(p-n)/d*s[d][c]+s[n][c];
		return ans+s[(p-n)%d][c];
	}
	return fib(p,c);
}
signed main()
{
	scanf("%s",a+1);
	n=strlen(a+1)/2;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<26;j++)s[i][j]=s[i-1][j];
		s[i][a[i]-'a']++;
	}
	KMP();
	l=read(),r=read();
	for(int i=0;i<26;i++)
		printf("%lld ",suan(r,i)-suan(l-1,i));
}