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 xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 200005;

LL l, r, ret[26], fib[100], cnt[100][26];
int n, f[MAXN];
char s[MAXN];

inline void Solve(LL r, int v)
{
	if (n % v == 0)
	{
		for (int i = 1; i <= n && i <= r; i ++)
			ret[s[i] - 'a'] ++;
		if (r >= n)
		{
			r -= n;
			for (int i = 1; i <= v; i ++)
				ret[s[i] - 'a'] += r / v + (r % v >= i);
		}
	}
	else
	{
		mset(cnt, 0);
		for (int i = 1; i <= v; i ++)
			cnt[0][s[i] - 'a'] ++;
		for (int i = 1; i <= n; i ++)
			cnt[1][s[i] - 'a'] ++;
		fib[0] = v, fib[1] = n;
		int i;
		for (i = 2; fib[i - 1] <= r; i ++)
		{
			fib[i] = fib[i - 1] + fib[i - 2];
			for (int j = 0; j < 26; j ++)
				cnt[i][j] = cnt[i - 1][j] + cnt[i - 2][j];
		}
		for (i --; i; i --)
			if (r >= fib[i])
			{
				for (int j = 0; j < 26; j ++)
					ret[j] += cnt[i][j];
				r -= fib[i];
			}
		for (int i = 1; i <= r; i ++)
			ret[s[i] - 'a'] ++;
	}
}

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	scanf("%s", s + 1); n = strlen(s + 1) >> 1;
	cin >> l >> r;
	f[1] = 0;
	for (int i = 2, j = 0; i <= n; j += s[i] == s[j + 1], f[i ++] = j)
		while (j && s[i] != s[j + 1])
			j = f[j];
	Solve(l - 1, n - f[n]);
	for (int i = 0; i < 26; i ++)
		ret[i] = -ret[i];
	Solve(r, n - f[n]);
	for (int i = 0; i < 26; i ++)
		printf("%lld%c", ret[i], i == 25 ? '\n' : ' ');
	return 0;
}