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<cstdio> 
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=2e6+5,M=1e7+50,mo=1e9+7;
int fail[N],n,i,per;
ll len1,len2,d,prt[30],l,r,f[305][26],g[305][26],sf[305],sg[305];
char s[N];
void kmp()
{
	int i,j;
	j=0;
	n=strlen(s+1)/2;
	fo(i,2,n)
	{
		while (j&&s[i]!=s[j+1]) j=fail[j];
		if (s[i]==s[j+1]) j++;
		fail[i]=j;
	}
}
void fib()
{
	int i,j;
	fo(i,2,300)
	{
		fo(j,0,25)  g[i][j]=g[i-1][j]+g[i-2][j];
		sg[i]=sg[i-1]+sg[i-2];
		if (sg[i]*2>r) break;
	}
	fo(i,0,300)
	{
		fo(j,0,25) f[i][j]=g[i][j]*2;
		sf[i]=sg[i]*2;
		if (sf[i]>r) break;
	}
}
void calc(ll x,ll xs)
{
	int pos;
	fo(pos,0,300)
		if (sg[pos]>x) break;
	while (pos>=0)
	{
		if (sg[pos]<=x)
		{
			fo(i,0,25) prt[i]+=xs*g[pos][i];
			x-=sg[pos];
			pos--;
		}else
		pos--;
	}
	fo(i,1,x) prt[s[i]-'a']+=xs; 
}
int main()
{
	scanf("%s %lld %lld",s+1,&l,&r);
	kmp();
	fo(i,1,n) g[0][s[i]-'a']++,sg[0]++;
	per=n-fail[n];
	fo(i,0,25) g[1][i]=g[0][i];
	sg[1]=sg[0];
	fo(i,1,per) g[1][s[i]-'a']++,sg[1]++;
	fib();
	calc(r,1);
	calc(l-1,-1);
	fo(i,0,25) printf("%lld ",prt[i]);
}