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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
const int maxn = 200005;

LL f[100][27];char s[maxn];int n,k,pre[maxn];
LL sum1[27],sum2[27];
void solve(LL n,int t,LL *c) {
	if (n<=0) return ;
	if (t<=2) {
		for (int j=1;j<=n;j++)
			c[s[j]-'a']++;
		return ;
	}
	if (f[t][26]==n) {
		for (int j=0;j<26;j++)
			c[j]+=f[t][j];
		return ;
	}
	if (f[t-1][26]>=n) solve(n,t-1,c);
	else {
		for (int j=0;j<26;j++)
			c[j]+=f[t-1][j];
		solve(n-f[t-1][26],t-2,c);
	}
}
int main()
{
	#ifdef Amberframe
		freopen("arc077f.in","r",stdin);
		freopen("arc077f.out","w",stdout);
	#endif
	scanf("%s",s+1);n=strlen(s+1)>>1;
	for (int i=2;i<=n;i++) {
		int p=pre[i-1];
		while (p&&s[p+1]!=s[i]) p=pre[p];
		if (s[p+1]==s[i]) pre[i]=p+1;
	}
	k=n-pre[n];f[1][26]=n;f[2][26]=n+k;
	for (int i=1;i<=n;i++) f[1][s[i]-'a']++;
	for (int i=1;i<=n+k;i++) f[2][s[i]-'a']++;
	int o=2;
	for (int i=3;;i++) {
		for (int j=0;j<=26;j++)
			f[i][j]=f[i-1][j]+f[i-2][j];
		o=i;
		if (f[i][26]>1e18) break;
	}
	
	LL l,r;scanf("%lld %lld",&l,&r);
	solve(r,o,sum1);solve(l-1,o,sum2);
	for (int i=0;i<26;i++)
		printf("%lld ",sum1[i]-sum2[i]);
	return 0;
}