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 <algorithm>
#include <cstring> 

using namespace std;

const int N = 2e5 + 10;

int n, T, p[N];
char s[N];
long long L, R, num[N][30];

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

long long calc(long long len, int c) {
    if (len <= n) return num[len][c];
    if (n % T == 0) {
        long long re = 1ll * (len - n) / T * num[T][c] + num[n][c];
        len -= n, len %= T;
        return re + num[len][c];
    }
    else return fib(len, c);
}

int main() {
    scanf("%s%lld%lld", s + 1, &L, &R);
    n = strlen(s + 1) / 2;
    p[1] = 0;
    for (int j = 0, i = 2; i <= n; i ++) {
        for (; j && s[j + 1] != s[i];) j = p[j];
        if (s[j + 1] == s[i]) j ++;
        p[i] = j;
    }
    for (int i = 1; i <= n; i ++) {
        for (int j = 0; j < 26; j ++) num[i][j] = num[i - 1][j];
        num[i][s[i] - 'a'] ++;
    }
    T = n - p[n];
	for (int i = 0; i < 26; i ++) printf("%lld%c", fib(R, i) - fib(L - 1, i), i == 25 ? 10 : 32);
    return 0;
}