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 <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef double LF;
typedef complex<LF> cd;

#define clr(x, y) memset(x, y, sizeof(x))
#define INF 0x7fffffff

const int MAXN = 2e5 + 15;

char ch[MAXN];
int n, nxt[MAXN], T;
LL l, r, f[MAXN][35];

inline int read(){
	int r = 0, z = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){if(ch == '-') z = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9'){r = r * 10 + ch - '0'; ch = getchar();}
	return r * z;
}

LL ff(LL len, int c){
	if(len <= n) return f[len][c];
	if(len <= n * 2) return f[n][c] + f[len - n][c];
	LL f1 = f[n][c], f2 = f[n][c] + f[T][c], l1 = n, l2 = n + T;
	while(len > l1 + l2){
		LL t = f2;
		f2 += f1; f1 = t;
		t = l2; l2 += l1; l1 = t;
	}
	return f2 + ff(len - l2, c);
}

LL calc(LL len, int c){
	if(len <= n) return f[len][c];
	if(len <= n * 2) return f[n][c] + f[len - n][c];
	if(n % T == 0){
		LL tmp = (len - 1ll * n) / T * f[T][c] + f[n][c];
		len -= n; len %= T;
		return tmp + f[len][c];
	}
	return ff(len, c);
}

void work(){
	scanf("%s %lld %lld", ch + 1, &l, &r); n = strlen(ch + 1) / 2;
	for(int i = 1; i <= n; i ++){
		for(int j = 0; j < 26; j ++) f[i][j] = f[i - 1][j];
		f[i][ch[i] - 'a'] ++;
	} nxt[1] = 0;
	for(int i = 2, j = 0; i <= n; i ++){
		for(; j && ch[j + 1] != ch[i];) j = nxt[j];
		if(ch[j + 1] == ch[i]) j ++;
		nxt[i] = j;
	}
	T = n - nxt[n];
	for(int i = 0; i < 26; i ++) printf("%lld ", calc(r, i) - calc(l - 1, i));
}

int main(){
	work();
	return 0;
}
