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 <iostream>
#include <string>
#define llint long long

using namespace std;

string s, t;
llint n, m;
llint l, r;
llint z[200005];
llint a[26], b[26];
llint len[105], num[105][26];

void z_algorithm(string &s, llint z[])
{
	z[0] = s.size();
	
	int x = 0, y = 0;
	for(int i = 1; i < s.size(); i++){
		if(i > y){
			z[i] = 0;
			for(int j = 0; j < s.size(); j++){
				if(i+j > s.size() || s[i+j] != s[j]) break;
				z[i]++;
			}
			x = i, y = i + z[i] - 1;
		}
		else if(i + z[i-x] <= y) z[i] = z[i-x];
		else{
			z[i] = y-i+1;
			for(int j = y-i+1; j < s.size(); j++){
				if(i+j > s.size() || s[i+j] != s[j]) break;
				z[i]++;
			}
			x = i, y = i + z[i] - 1;
		}
	}
}

void calc(llint x, llint a[])
{
	if(x < 0) return;
	for(int i = 0; i < n; i++) a[s[i]-'a'] += x/n;
	for(int i = 0; i < x%n;i++) a[s[i]-'a']++;
}

void calc2(llint x, llint a[])
{
	if(x <= n){
		for(int i = 0; i < x; i++) a[s[i]-'a']++;
		return;
	}
	for(int i = 2; ; i++){
		if(x < len[i]){
			for(int j = 0; j < 26; j++) a[j] += num[i-1][j];
			calc2(x-len[i-1], a);
			break;
		}
	}
}

int main(void)
{
	cin >> s >> l >> r;
	s = s.substr(0, (int)s.size()/2);
	n = s.size();
	
	z_algorithm(s, z);
	int p = -1;
	for(int i = 1; i < n; i++){
		if(z[i] == n-i){
			p = i;
			break;
		}
	}
	if(p == -1){
		calc(r, a), calc(l-1, b);
		for(int i = 0; i < 26; i++) cout << a[i]-b[i] << " "; cout << endl;
		return 0;
	}
	
	t = s.substr(0, p);
	m = t.size();
	
	len[0] = m, len[1] = n;
	for(int i = 0; i < m; i++) num[0][t[i]-'a']++;
	for(int i = 0; i < n; i++) num[1][s[i]-'a']++;
	for(int i = 2; ; i++){
		len[i] = len[i-1] + len[i-2];
		for(int j = 0; j < 26; j++) num[i][j] = num[i-1][j] + num[i-2][j];
		if(len[i] > 1000000000000000000) break;
	}
	
	calc2(r, a), calc2(l-1, b);
	for(int i = 0; i < 26; i++) cout << a[i]-b[i] << " "; cout << endl;
	return 0;
}