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 <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <ctime>
 
#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)
 
using namespace std;
 
typedef long long LL;
const int N = 400010;
 
char s0[N];
int len, nxt[N];
 
void get_nxt()
{
	int j = 0;
	for (int i = 2; i <= len; i ++) {
		while (j && s0[i] != s0[j + 1]) j = nxt[j];
		if (s0[i] == s0[j + 1]) j ++;
		nxt[i] = j; 
	}
}
 
LL f[N][30], ans[30];
 
LL calc(int x, LL k, int c)
{
	if (x == 1 || x == 0) {
		int ret = 0;
		Rep(i, k) if (s0[i] - 'a' + 1 == c) ret ++;
		return ret;
	}
	
	if (f[x][0] == k) return f[x][c];
	if (k >= f[x - 1][0])
		return f[x - 1][c] + calc(x - 2, k - f[x - 1][0], c);
	else return calc(x - 1, k, c);
}
 
int main()
{
	scanf("%s", s0 + 1);
	len = strlen(s0 + 1) / 2;
	LL l, r;
	scanf("%lld%lld", &l, &r);
	get_nxt();
	
	Rep(i, len - nxt[len]) f[0][s0[i] - 'a' + 1] ++;
	Rep(i, len) f[1][s0[i] - 'a' + 1] ++;
	f[0][0] = len - nxt[len], f[1][0] = len;
	int x = 2;
	while (true) {
		Rep0(i, 26) f[x][i] = f[x - 1][i] + f[x - 2][i];
		if (f[x][0] >= r) break;
		x ++;
	}

	if (l > 1) Rep0(i, x) if (f[i][0] >= l - 1) {
		Rep(j, 26) ans[j] -= calc(i, l - 1, j);
		break;
	}
 
	Rep0(i, x) if (f[i][0] >= r) {
		Rep(j, 26) ans[j] += calc(i, r, j);
		break;
	}
	Rep(i, 26) printf("%lld ", ans[i]);
	
	return 0;
}