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
#include<bits/stdc++.h>
#define rep(i,a,b) for (int i=(a); i<=(b); i++)
#define per(i,a,b) for (int i=(a); i>=(b); i--)
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1; char ch=getchar();
    while (!(ch>='0'&&ch<='9')) {if (ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+(ch-'0'); ch=getchar();}
    return x*f;
}

const int N = 200005;
const int M = 105;
const LL INF = 1000000000000000000;

int n,pre[N];
int num[26][N];
int tot=1;
LL sum1[M],sum2[26][M];
int T;
char s[N];
LL l,r;

LL Calc(int c,LL lim) {
	LL ret=0;
	per(i,tot,1) if (lim>=sum1[i]) lim-=sum1[i],ret+=sum2[c][i];
	return ret+num[c][lim];
}

int main() {

	#ifndef ONLINE_JUDGE
	//	freopen("data.in","r",stdin);
	//	freopen("data.out","w",stdout);
	#endif

	scanf("%s",s+1); scanf("%lld%lld",&l,&r);
	n=strlen(s+1)/2;
	
	pre[1]=0; int k=0;
	rep(i,2,n) {
		while (k&&s[k+1]!=s[i]) k=pre[k];
		if (s[k+1]==s[i]) k++;
		pre[i]=k;
	} T=n-pre[n];
	
	rep(i,1,n) {
		rep(j,0,25) num[j][i]=num[j][i-1];
		num[s[i]-'a'][i]++;
	}
	
	sum1[0]=T; sum1[1]=n; rep(i,0,25) sum2[i][0]=num[i][T],sum2[i][1]=num[i][n];
	while (sum1[tot]<=r) {++tot; sum1[tot]=sum1[tot-1]+sum1[tot-2]; rep(i,0,25) sum2[i][tot]=sum2[i][tot-1]+sum2[i][tot-2];}
	
	rep(i,0,25) {
		printf("%lld",Calc(i,r)-Calc(i,l-1));
		if (i!=25) putchar(' '); else puts("");
	}

	return 0;
}
