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 LL long long

const int maxn=2e5+10;

int n,next[maxn];
LL L,R,c0[30],cA[30],cB[30],Ans[30],LA,LB;
char s[maxn];

void KMP()
{
	int i;
	next[0]=0; next[1]=0;
	for (i=2;i<=n;i++)
	{
		next[i]=next[i-1];
		while (s[next[i]+1]!=s[i]&&next[i])
			next[i]=next[next[i]];
		next[i]+=s[next[i]+1]==s[i];
	}
}

void solve(LL R,int flag)
{
	int i;
	if (R<=n)
	{
		for (i=1;i<=R;i++)
			Ans[s[i]-'a']+=flag;
		return;
	}
	LA=next[n];
	LB=n-2*next[n];
	memset(cA,0,sizeof(cA));
	memset(cB,0,sizeof(cB));
	for (i=1;i<=LA;i++) cA[s[i]-'a']++;
	for (i=LA+1;i<=LA+LB;i++) cB[s[i]-'a']++;
	while (LA+LB+LA+LA+LB<=R)
	{
		swap(LA,LB); LA+=LB;
		for (i=0;i<26;i++)
			swap(cA[i],cB[i]),cA[i]+=cB[i];
	}
	for (i=0;i<26;i++)
		Ans[i]+=flag*(cA[i]+cB[i]+cA[i]);
	R-=LA+LB+LA;
	while (R)
	{
		if (LA+LB+LA==n) break;
		LA-=LB; swap(LA,LB);
		for (i=0;i<26;i++)
			cA[i]-=cB[i],swap(cA[i],cB[i]);
		if (LA+LB+LA==n) break;
		if (R>=LA)
		{
			for (i=0;i<26;i++)
				Ans[i]+=flag*cA[i];
			R-=LA;
		}
	}
	for (i=1;i<=R;i++)
		Ans[s[i]-'a']+=flag;
}

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i;
	scanf("%s",s+1);
	scanf("%lld%lld",&L,&R);
	n=strlen(s+1);
	n=n/2;
	KMP();
	if (n%(n-next[n])==0)
	{
		for (i=1;i<=n;i++) c0[s[i]-'a']++;
		for (i=0;i<26;i++) Ans[i]+=c0[i]*((R/n)-((L-1)/n));
		for (i=1;i<=R%n;i++) Ans[s[i]-'a']++;
		for (i=1;i<=(L-1)%n;i++) Ans[s[i]-'a']--;
		for (i=0;i<26;i++) printf("%lld ",Ans[i]);
		return 0;
	}
	while (next[n]>=(n+1)/2)
		n+=n-next[n],KMP();
	solve(R,+1);
	solve(L-1,-1);
	for (i=0;i<26;i++)
		printf("%lld ",Ans[i]);
}