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 <fstream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cassert>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
#include <queue>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <stack>
#include <iomanip>
#include <utility>
#include <functional>
using namespace std;
#define ll long long
#define ull unsigned ll
int n, fail[100005];
int len;
ll pre[100005][26];
ll l, r;
string s;
int gaofail(string &str) {
	fail[0]=0;
	fail[1]=0;
	for (int i=1, j=0; i<n; i++) {
		while (j&&str[i]!=str[j]) j=fail[j];
		if (str[i]==str[j]) j++;
		fail[i+1]=j;
	}
	return fail[n];
}
void gaopre() {
	pre[0][s[0]-'a']=1ll;
	for (int i=1; i<n; i++) {
		for (int j=0; j<26; j++) pre[i][j]=pre[i-1][j];
		pre[i][s[i]-'a']++;
	}
}
ll calc1(ll i, int c) {
	if (i<0)
		return 0;
	if (i<n)
		return pre[i][c];
	i-=n;
	return pre[n-1][c]+pre[len-1][c]*(i/len)+pre[i%len][c];
}
ll calc2(ll i, int c) {
	if (i<0)
		return 0;
	if (i<n)
		return pre[i][c];
	if (i<2*n)
		return pre[n-1][c]+pre[i-n][c];
	ll f1=n, f2=n+len, r1=pre[n-1][c], r2=pre[n-1][c]+pre[len-1][c];
	while (f1+f2<i) {
		ll t=f2;
		f2+=f1;
		f1=t;
		t=r2;
		r2+=r1;
		r1=t;
	}
	return r2+calc2(i-f2, c);
}
int main() {
	cin>>s;
	cin>>l>>r;
	l--, r--;
	s=s.substr(0, s.length()/2);
	n=s.length();
	len=n-gaofail(s);
	gaopre();
	if (n%len==0) {
		for (int i=0; i<26; i++)
			cout<<calc1(r, i)-calc1(l-1, i)<<(i<25?' ':'\n');
	} else {
		for (int i=0; i<26; i++)
			cout<<calc2(r, i)-calc2(l-1, i)<<(i<25?' ':'\n');
	}
	return 0;
}