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, x, y) for(int i = (x), _ = (y); i <= _; ++ i)
#define per(i, x, y) for(int i = (x), _ = (y); i >= _; -- i)
#define dprintf(...) fprintf(stderr, __VA_ARGS__)
#define disp(x) cout << #x << " = " << x << "; "

typedef long long LL;

template <class T> bool chkmin(T& a, T b) { return a > b ? a = b, true : false; }
template <class T> bool chkmax(T& a, T b) { return a < b ? a = b, true : false; }

template <class T> void read(T& a) {
	char c = getchar(); 
	T f = 1; 
	a = 0;
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) a = a * 10 + c - '0';
	a *= f;
}
/*}}}*/

const int maxN = 2e5 + 5;

char s[maxN];
LL fib[100];
int fail[maxN];
int lens, lent;
LL ansl[26], ansr[26]; 
int sums[26], sumt[26];
LL L, R;

void getans(LL Bnd, LL ans[])
{
	if(Bnd < lens)
	{
		rep(i, 1, Bnd) ans[s[i] - 'a']++;
		return;
	}
	rep(i, 0, 90)
	{
		if(fib[i] * lent + fib[i + 1] * lens > Bnd)
		{
			rep(j, 0, 25) ans[j] += fib[i - 1] * sumt[j] + fib[i] * sums[j];
			getans(Bnd - fib[i - 1] * lent - fib[i] * lens, ans);
			return;
		}
	}
}

int main()
{
#ifdef Leeson
	freopen("tmp.in", "r", stdin);
	freopen("tmp.out", "w", stdout);
#endif

	fib[0] = 0; fib[1] = 1;
	rep(i, 2, 90) fib[i] = fib[i - 1] + fib[i - 2];
	scanf("%s", s + 1);
	lens = strlen(s + 1) >> 1;
	fail[0] = fail[1] = 0;
	rep(i, 2, lens)
	{
		int tmp = fail[i - 1];
		while(tmp && s[tmp + 1] != s[i]) tmp = fail[tmp];
		fail[i] = s[tmp + 1] == s[i] ? tmp + 1 : 0;
	}
	lent = lens - fail[lens];
	rep(i, 1, lent) sumt[s[i] - 'a']++;
	rep(i, 1, lens) sums[s[i] - 'a']++;
	read(L), read(R);
	getans(R, ansr);
	getans(L-1, ansl);
	rep(i, 0, 25) printf("%lld ", ansr[i] - ansl[i]);

	return 0;
}