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>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 2e5 + 10;
const int M = 110;

typedef long long LL;

char S[N];
int n, f[N];

struct Info{
	LL cnt[26];

	Info(){
		memset(cnt, 0, sizeof cnt);
	}

	void operator = (const Info& A){
		For(i, 0, 25) cnt[i] = A.cnt[i];
	}

	Info operator + (const Info& A) const{
		Info ret;
		For(i, 0, 25) ret.cnt[i] = A.cnt[i] + cnt[i];
		return ret;
	}

	Info operator - (const Info& A) const{
		Info ret;
		For(i, 0, 25) ret.cnt[i] = cnt[i] - A.cnt[i];
		return ret;
	}

}A[M];

LL len[M];

Info work(int x, LL r){
	if(x <= 2){
		Info ret;
		For(i, 1, r) ret.cnt[S[i] - 'a']++;
		return ret;
	}
	if(r < len[x - 1]) return work(x - 1, r);
	else return A[x - 1] + work(x - 1, r - len[x - 1]);
}

Info calc(LL r){
	if(r <= n) return work(1, r);
	int i = 1;
	while(len[i] * 2 < r) ++i;
	return A[i] + work(i, r - len[i]);
}

int main(){
	scanf("%s", S + 1);
	n = strlen(S + 1) / 2;

	f[1] = 0;
	For(i, 2, n){
		int p = f[i - 1];
		while(p && S[p + 1] != S[i]) p = f[p];
		f[i] = S[p + 1] == S[i] ? p + 1 : 0;
	}
//	printf("period = %d\n", n - f[n]);

	For(i, 1, n) A[1].cnt[S[i] - 'a']++;
	len[1] = n;
	For(i, 1, n + n - f[n]) A[2].cnt[S[i] - 'a']++;
	len[2] = n + n - f[n];
	For(i, 3, 100) len[i] = len[i - 1] + len[i - 2], A[i] = A[i - 1] + A[i - 2];

	LL l, r;
	scanf("%lld%lld", &l, &r);
	Info ans = calc(r) - calc(l - 1);
	For(i, 0, 25) printf("%lld%c", ans.cnt[i], i == 25 ? '\n' : ' ');
	return 0;
}