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>
using namespace std;
const int N = 2e5 + 9;

char s[N];
int f[N], n, t, lim;
using ll = long long;
ll l, r, c[120][26], ans[26], fib[N];

void Solve (ll x, int t) {
	ll bs = x / ::t, b2 = x % ::t;
	bs *= t;
	for (int i = 0; i < ::t; ++i) {
		ans[s[i] - 'a'] += bs;
		if (i < b2) ans[s[i] - 'a'] += t;
	}
}

void Solve2 (ll x, int t) {
	int now = lim + 1;
	while (now-- > 0) {
		if (x >= fib[now]) {
			for (int i = 0; i < 26; ++i) ans[i] += c[now][i] * t;
			x -= fib[now], --now;
		}
	}
	for (int i = 0; i < x; ++i) ans[s[i] - 'a'] += t;
}

int main () {
	scanf("%s", s);
	n = strlen(s);
	n >>= 1;
	s[n] = '\0';
	scanf("%lld%lld", &l, &r);
	for (int i = 1, j; i < n; ++i) {
		j = f[i];
		while (j && s[j] != s[i]) j = f[j];
		f[i + 1] = s[i] == s[j] ? j + 1 : 0;
	}
	t = n - f[n];
	if (n % t == 0) {
		Solve(r, 1); Solve(l - 1, -1);
	} else {
		for (int i = 0; i < t; ++i) ++c[0][s[i] - 'a'];
		for (int i = 0; i < n; ++i) ++c[1][s[i] - 'a'];
		fib[0] = t; fib[1] = n;
		for (lim = 2; ; ++lim) {
			for (int j = 0; j < 26; ++j) c[lim][j] = c[lim - 1][j] + c[lim - 2][j];
			if ((fib[lim] = fib[lim - 1] + fib[lim - 2]) >= 1e18) break;
		}
		Solve2(r, 1); Solve2(l - 1, -1);
	}
	ll sum = 0;
	for (int i = 0; i < 26; ++i) printf("%lld ", ans[i]), sum += ans[i];
	return 0;
}