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;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

vector<int> mp(string S){
	vector<int> V(S.size()+1,-1);
	int j = -1;
	repp(i,0,S.size()){
		while(j >= 0 && S[i] != S[j]) j = V[j];
		++j;
		V[i+1] = j;
	}
	return V;
}

vector<LL> solve(LL z, vector<vector<LL>> &A, string S[2], int p){
	vector<LL> V(26,0);
	if(z <= 0) return V;
	if(z >= A[p][26]){
		repp(i,0,26) V[i] = A[p][i];
	} else if(p > 1){
		V = solve(z,A,S,p-1);
		auto W = solve(z-A[p-1][26],A,S,p-2);
		repp(i,0,26) V[i] += W[i];
	} else {
		repp(i,0,z) ++V[S[p][i]-'a'];
	}
	return V;
}

int main(){
	string S; cin >> S;
	LL l,r; cin >> l >> r;
	string T = S.substr(0,S.size()/2);
	int x = mp(T)[T.size()];
	vector<vector<LL>> A(2,vector<LL>(27,0));
	string U[2];
	if(x*2 <= T.size()){
		U[0] = T.substr(x,T.size()-x*2);
		U[1] = T.substr(0,x);
	} else {
		U[0] = T.substr(0,x);
		U[1] = T.substr(0,T.size()-x);
	}
	repp(k,0,2) repp(i,0,U[k].size()){
		++A[k][U[k][i]-'a'];
		++A[k][26];
	}
	x = 1;
	while(A[x][26] < r){
		A.push_back(vector<LL>(27));
		repp(i,0,27) A[x+1][i] = A[x][i] + A[x-1][i];
		++x;
	}
	auto B = solve(r,A,U,x);
	auto C = solve(l-1,A,U,x);
	repp(i,0,26) cout << B[i]-C[i] << (i==25?'\n':' ');
	return 0;
}