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 <cstring>
#include <algorithm>

using namespace std;

const int N=2e5+5;

typedef long long LL;

char S[N],T[N];

int n,nn,m,Fail[N],ss[N],cs[N],ct[N];

LL l,r,cnt[N],fs[N],ft[N],g[N];

void calc(LL t,int sig)
{
	if (!t) return;
	memset(ss,0,sizeof(ss));
	for (int i=m-1;i>=0;i--) if (t>=g[i])
	{
		t-=g[i]; ss[i]++;
	}
	for (int i=m-1;i>=0;i--) if (ss[i])
	{
		for (int c=0;c<26;c++)
		{
			cnt[c]+=sig*(fs[i]*cs[c]+ft[i]*ct[c]);
		}
	}
	if (t)
	{
		for (int i=1;i<=t;i++) cnt[S[i]-'a']+=sig;
	}
}

int main()
{
	scanf("%s%lld%lld",S+1,&l,&r); n=strlen(S+1)/2;
	for (int i=2,j=0;i<=n;i++)
	{
		for (;j>0 && S[i]!=S[j+1];j=Fail[j]);
		if (S[i]==S[j+1]) j++;
		Fail[i]=j;
	}
	for (int i=1;i<=n-Fail[n];i++) T[++nn]=S[i];
	for (int i=1;i<=n;i++) cs[S[i]-'a']++;
	for (int i=1;i<=nn;i++) ct[T[i]-'a']++;
	fs[0]=fs[1]=ft[1]=1; g[0]=n; g[1]=nn+n;
	for (m=2;g[m-1]<r;m++) fs[m]=fs[m-1]+fs[m-2],ft[m]=ft[m-1]+ft[m-2],g[m]=g[m-1]+g[m-2];
	if (n>=r) calc(r,1);else
	{
	for (int i=0;i<m;i++) if (g[i+1]>=(r+1)/2)
	{
		calc(g[i],1); calc(r-g[i],1); break;
	}
	}
	if (n>=l-1) calc(l-1,-1);else
	{
	for (int i=0;i<m;i++) if (g[i+1]>=l/2)
	{
		calc(g[i],-1); calc(l-1-g[i],-1); break;
	}
	}
	for (int c=0;c<26;c++) printf("%lld ",cnt[c]);
    return 0;
}