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>
#define ll long long

using namespace std;

const int N = 2e5 + 5;

ll l, r;
int len, S, T;
int ans[30][N], nxt[N];
char A[N];

ll cal(int a, ll now){
    if (now <= len) return ans[a][now];
    if (now <= 2 * len) return ans[a][len] + ans[a][now - len];
    ll f1 = ans[a][S], f2 = ans[a][len], lenf1 = S, lenf2 = len;
    while (now > lenf1 + lenf2) {
        ll tem = f2, lentem = lenf2;
        f2 += f1; lenf2 += lenf1;
        f1 = tem; lenf1 = lentem;
    }
    return f2 + cal(a, now - lenf2);
}

int main(){
    scanf("%s", A + 1);
    scanf("%lld%lld", &l, &r);
    len = strlen(A + 1) / 2;
    
    nxt[1] = 0;
    for (int i = 2; i <= len; i ++) {
        int j = nxt[i - 1];
        while(j != 0 && A[j + 1] != A[i]) j = nxt[j];

        if(A[j + 1] == A[i]) nxt[i] = j + 1; 
        else nxt[i] = 0;
    }
    T = nxt[len];
    S = len - T;
    
    for (int i = 1; i <= 26; i ++) {
        for (int j = 1; j <= len; j ++) {
            ans[i][j] = ans[i][j - 1];
            if ((int)A[j] - 96 == i) ans[i][j] ++;
        }
    }
    for (int i = 1; i <= 26; i ++)
        printf("%lld ", cal(i, r) - cal(i, l - 1));
    printf("\n");
    return 0;
}