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 <bits/stdc++.h>

using namespace std;

# define REP(i, a, b) for(int i = a; i <= b; ++ i)
# define REPD(i, a, b) for(int i = a; i >= b; -- i)
# define CLR(i, a) memset(i, a, sizeof(i))
# define REPG(i, h, x) for(int i = h[x]; ~i; i = edge[i].next)

typedef long long LL;
const int N = 2e5 + 3;

char buf[N];
LL suc[N][30];
int n;

int pe, nxt[N];
LL fib(LL len, int c) {
    if(len <= n) return suc[len][c];
    if(len <= n * 2) return suc[n][c] + suc[len - n][c];
    LL g1 = suc[n][c], g2 = suc[n][c] + suc[pe][c], l1 = n, l2 = n + pe;
    while(len > l1 + l2) {
        LL tg = g2; g2 += g1, g1 = tg;
        LL tl = l2; l2 += l1, l1 = tl;
    }
    return g2 + fib(len - l2, c);
}
LL lf, rt;
LL calc(LL len, int c) {
    if(len <= n) return suc[len][c];
    else if(n % pe) return fib(len, c);
    else {
        LL ret = 1ll * (len - n) / pe * suc[pe][c] + suc[n][c];
        len -= n, len %= pe;
        return ret + suc[len][c];
    }
}

inline void get_nxt() {
    int j = 0; nxt[1] = 0;
    REP(i, 2, n) {
        while(j && buf[j + 1] != buf[i]) j = nxt[j];
        if(buf[j + 1] == buf[i]) ++ j;
        nxt[i] = j;
    }

}

int main() {
    scanf("%s%lld%lld", buf + 1, &lf, &rt);
    n = strlen(buf + 1); n >>= 1;
    get_nxt();
    for(int i = 1; i <= n; i ++) {
        for(int j = 0; j < 26; j ++) suc[i][j] = suc[i - 1][j];
        suc[i][buf[i] - 'a'] ++;
    }
    pe = n - nxt[n];
    REP(i, 0, 25) printf("%lld ", calc(rt, i) - calc(lf - 1, i));
    return 0;
}