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>
using namespace std;
#define MAXN	200005
#define MAXC	26
#define MAXLOG	105
#define INF	1e18
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
long long cnt[MAXLOG][MAXC];
long long sum[MAXLOG];
long long ans[MAXC], bns[MAXC];
long long n, l, r, m;
char s[MAXN];
void init() {
	static int next[MAXN];
	for (int i = 2; i <= n; i++) {
		int j = next[i - 1];
		while (j && s[j + 1] != s[i]) j = next[j];
		if (s[j + 1] == s[i]) j++;
		next[i] = j;
	}
	for (int i = next[n] + 1; i <= n; i++)
		cnt[1][s[i] - 'a']++;
	m = 1;
	sum[0] = n;
	for (int i = 1; i <= n; i++)
		cnt[0][s[i] - 'a']++;
	sum[1] = n + n - next[n];
	for (int i = 0; i < 26; i++)
		cnt[1][i] += cnt[0][i];
	while (sum[m] + sum[m - 1] <= INF) {
		m++; sum[m] = sum[m - 1] + sum[m - 2];
		for (int i = 0; i < 26; i++)
			cnt[m][i] = cnt[m - 1][i] + cnt[m - 2][i];
	}
}
void calc(long long total, long long *ans) {
	for (int i = m; i >= 0; i--) {
		if (total < sum[i]) continue;
		total -= sum[i];
		for (int j = 0; j < 26; j++)
			ans[j] += cnt[i][j];
	}
	for (int i = 1; i <= total; i++)
		ans[s[i] - 'a']++;
}
int main() {
	scanf("%s", s + 1);
	n = strlen(s + 1) / 2;
	read(l), read(r);
	init();
	calc(r, ans);
	calc(l - 1, bns);
	for (int i = 0; i < 26; i++)
		printf("%lld ", ans[i] - bns[i]);
	return 0;
}