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>
typedef long long int64;
static const int MAXN = 2e5 + 4;
static const int LOGL = 90;

int n, k;   // k = minimum period
char s[MAXN];
int pre[MAXN];
int s_sum[MAXN][26] = {{ 0 }};

int64 fib[LOGL];

int64 l, r;
int64 ans[26] = { 0 };

void count(int64 r, int mul)
{
    if (r <= n) {
        for (int i = 0; i < 26; ++i) ans[i] += s_sum[r][i] * mul;
    } else {
        int i;
        for (i = 0; fib[i + 1] * n + fib[i] * k <= r; ++i) ;
        --i;
        //printf("R = %lld | S * %lld + T * %lld + %lld\n", r, fib[i + 1], fib[i], r - (fib[i + 1] * n + fib[i] * k));
        for (int j = 0; j < 26; ++j)
            ans[j] += (fib[i + 1] * s_sum[n][j] + fib[i] * s_sum[k][j]) * mul;
        r -= (fib[i + 1] * n + fib[i] * k);
        count(r, mul);
    }
}

int main()
{
    scanf("%s", s);
    n = strlen(s) / 2;

    pre[0] = -1; pre[1] = 0;
    int last = 0;
    for (int i = 2; i <= n; ++i) {
        while (last > 0 && s[last] != s[i - 1]) last = pre[last];
        if (s[last] != s[i - 1]) pre[i] = 0;
        else pre[i] = ++last;
    }
    k = n - pre[n];

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < 26; ++j)
            s_sum[i + 1][j] = s_sum[i][j] + (s[i] == j + 'a');
    }

    fib[0] = 0; fib[1] = 1;
    for (int i = 2; i < LOGL; ++i) fib[i] = fib[i - 1] + fib[i - 2];

    scanf("%lld%lld", &l, &r);

    count(r, +1);
    count(l - 1, -1);
    for (int i = 0; i < 26; ++i) printf("%lld%c", ans[i], i == 25 ? '\n' : ' ');

    return 0;
}
