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>
using namespace std;
typedef long long ll;
const int N=200200,D=26;
char s[N];
int p[N];
ll len[1000],num[1000];
int i,j,k,n,m;
ll l,r;
ll work(ll x,int ch) {
	int i,j;
	ll ans=0;
	ch+='a';
	len[1]=n;num[1]=0;
	for (i=1;i<=n;i++) if (s[i]==ch) num[1]++;
	len[2]=n+m;num[2]=num[1];
	for (i=1;i<=m;i++) if (s[i]==ch) num[2]++;
	//printf("%lld %lld\n",len[1],len[2]);
	for (i=2;i;i++) {
		//printf("%lld\n",len[i]);
		if (len[i-1]+len[i]>x) break;
		len[i+1]=len[i-1]+len[i];
		num[i+1]=num[i-1]+num[i];
	}
	for (j=i;j;j--) {
		if (x>=len[j]) {
			x-=len[j];
			ans+=num[j];
		}
	}
	for (i=1;i<=x;i++) if (s[i]==ch) ans++;
	return ans;
}
int main() {
	scanf("%s",s+1);
	n=strlen(s+1);
	n/=2;
	j=0;
	for (i=2;i<=n;i++) {
		while (j && s[j+1]!=s[i]) j=p[j];
		if (s[j+1]==s[i]) p[i]=++j;
	}
	m=n-p[n];
	scanf("%lld%lld",&l,&r);
	for (i=0;i<D;i++) printf("%lld ",work(r,i)-work(l-1,i));
	puts("");
}