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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <algorithm>
using namespace std;

#define Rep(i,l,r) for (int i = (l); i <= (r); ++ i)
#define RepD(i,r,l) for (int i = (r); i >= (l); -- i)
#define RepG(i,x) for (int i = hd[x]; i; i = e[i].nxt)

typedef long long ll;
const int N = 4e5+5;
int n, bd, f[N][35], ne[N]; ll l, r;
char s[N];

void pre () {
	Rep(i,1,n) {
		++ f[i][s[i]-'a'+1];
		Rep(j,1,26) f[i][j] += f[i-1][j];
	}
	int t = 2, k = ne[1] = 0;
	while (t<=n) {
		while (k && s[k+1]!=s[t]) k = ne[k];
		if (s[k+1]==s[t]) ++ k; ne[t++] = k;
	}
	bd = n-ne[n];
}

ll calc (ll k, int c) {
	if (k<=n) return f[k][c];
	if (k<=(n<<1)) return f[n][c]+f[k-n][c];
	ll lsf = f[n][c], nwf = f[n][c]+f[bd][c], lsl = n, nwl = n+bd;
	while (k>lsl+nwl) {
		swap (lsl, nwl); nwl = lsl+nwl;
		swap (lsf, nwf); nwf = lsf+nwf;
	}
	return nwf+calc (k-nwl, c);
}

int main () {
	scanf ("%s%lld%lld", s+1, &l, &r);
	n = strlen (s+1)>>1; pre ();
	Rep(i,1,26) printf ("%lld ", calc (r,i)-calc (l-1,i));
	puts ("");
	return 0;
}