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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;

const int maxn = 200009;
char s[maxn]; int Next[maxn], n, m;
LL l, r, w[26];

void solve(LL k){
	if (k <= n){
		for (int i=0; i<k; i++) w[s[i]-'a']++;
		return;
	}
	LL S = 1, T = 0, _s = 0, _t = 1;
	while (S*n + T*m < k) { S += _s; T += _t; _s = S - _s; _t = T - _t; }
	solve(k - _s*n - _t*m);
	for (int i=0; i<n; i++) w[s[i]-'a'] += _s;
	for (int i=0; i<m; i++) w[s[i]-'a'] += _t;
}

int main(){
	scanf("%s", &s); scanf("%lld%lld", &l, &r);
	n = strlen(s)/2;
	int k = 0; Next[0] = 0;
	for (int i=1; i<n; i++){
		while (k && s[k] != s[i]) k = Next[k-1];
		if (s[k] == s[i]) k++; Next[i] = k;
	}
	m = n-Next[n-1];
	solve(l-1); for (int i=0; i<26; i++) w[i] = -w[i];
	solve(r); for (int i=0; i<26; i++) printf("%lld ", w[i]);
	return 0;
}