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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

const int N=2*1e5+5;

ll l,r,L[N],cnt[N][26];
int n,tot,nxt[N];
char st[N];

void kmp() {
	int k=0;
	fo(i,2,n) {
		while (k&&st[k+1]!=st[i]) k=nxt[k];
		nxt[i]=k+=st[k+1]==st[i];
	}
}

ll calc(ll n,int c) {
	int id=0;
	fd(i,tot,1) if (L[i]<=n) {id=i;break;}
	if (!id) {
		ll res=0;
		fo(i,1,n) if (st[i]-'a'==c) res++;
		return res;
	}
	return cnt[id][c]+calc(n-L[id],c);
}

int main() {
	scanf("%s",st+1);n=strlen(st+1);
	n/=2;kmp();
	L[0]=n-nxt[n];fo(i,1,n-nxt[n]) cnt[0][st[i]-'a']++;
	L[1]=n;fo(i,1,n) cnt[1][st[i]-'a']++;
	scanf("%lld%lld",&l,&r);
	for(tot=1;L[tot]<r;) {
		tot++;
		L[tot]=L[tot-1]+L[tot-2];
		fo(j,0,25) cnt[tot][j]=cnt[tot-1][j]+cnt[tot-2][j];
	}
	fo(c,0,25) printf("%lld ",calc(r,c)-calc(l-1,c));
	return 0;
}