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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

const int MAXN = 200000;

ll a[26], b[26], c[150][26], d[150];

char S[MAXN + 5]; int f[MAXN + 5];
int main() {
	int n, m; ll l, r;
	scanf("%s%lld%lld", S + 1, &l, &r);
	n = strlen(S + 1), m = n / 2;
	
	f[0] = -1, f[1] = 0;
	for(int i=2;i<=m;i++) {
		int j = f[i - 1];
		while( j != -1 && S[i] != S[j + 1] )
			j = f[j];
		f[i] = j + 1;
	}
	
	int p = m - f[m];
	if( f[m] && m % (m - f[m]) == 0 ) {
		ll q1 = r / p, r1 = r % p, q2 = (l - 1) / p, r2 = (l - 1) % p;
		for(int i=1;i<=p;i++) b[S[i] - 'a']++;
		for(int i=0;i<26;i++) a[i] += q1 * b[i];
		for(int i=1;i<=r1;i++) a[S[i] - 'a']++;
		for(int i=0;i<26;i++) a[i] -= q2 * b[i];
		for(int i=1;i<=r2;i++) a[S[i] - 'a']--;
	}
	else {
		for(int i=1;i<=p;i++) c[0][S[i] - 'a']++;
		for(int i=1;i<=m;i++) c[1][S[i] - 'a']++;
		int k; d[0] = p, d[1] = m;
		for(k=1;d[k]<=r;k++) {
			for(int j=0;j<26;j++)
				c[k + 1][j] = c[k][j] + c[k - 1][j];
			d[k + 1] = d[k] + d[k - 1];
		}
		
		ll t = r;
		for(int i=k;i>=0;i--) {
			if( t >= d[i] ) {
				for(int j=0;j<26;j++)
					a[j] += c[i][j];
				t -= d[i];
			}
		}
		for(int i=1;i<=t;i++)
			a[S[i] - 'a']++;
		
		t = l - 1;
		for(int i=k;i>=0;i--) {
			if( t >= d[i] ) {
				for(int j=0;j<26;j++)
					a[j] -= c[i][j];
				t -= d[i];
			}
		}
		for(int i=1;i<=t;i++)
			a[S[i] - 'a']--;
	}
	
	for(int i=0;i<26;i++) printf("%lld ", a[i]);
}