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
#include <iostream>
#include<vector>
#include<cstring>
using namespace std;

typedef long long lint;
const int MAXN = 200005;

char str[MAXN];
int fail[MAXN];
lint dp[222];
lint cnt[222][26];

vector<lint> solve(lint p, int s, int t){
	memset(dp, 0, sizeof(dp));
	memset(cnt, 0, sizeof(cnt));
	vector<lint> dap(26);
	dp[0] = s;
	dp[1] = s+t;
	vector<int> v;
	for(int i=0; i<s; i++){
		cnt[0][str[i] - 'a']++;
	}
	for(int i=0; i<s+t; i++){
		cnt[1][str[i] - 'a']++;
	}
	for(int i=2; i<100; i++){
		for(int j=0; j<26; j++){
			cnt[i][j] = cnt[i-1][j] + cnt[i-2][j];
		}
		dp[i] = (dp[i-1] + dp[i-2]);
		if(dp[i] >= p){
			for(int j=i; j>=0; j--){
				while(p >= dp[j]){
					p -= dp[j];
					v.push_back(j);
				}
			}
			break;
		}
	}
	for(int i=0; i<p; i++){
		dap[str[i] - 'a']++;
	}
	for(auto &i : v){
		for(int j=0; j<26; j++){
			dap[j] += cnt[i][j];
		}
	}
	return dap;
}

int main(){
	scanf("%s", str);
	int p = 0;
	for(int i=1; str[i]; i++){
		while(p && str[i] != str[p]) p = fail[p];
		if(str[i] == str[p]) p++;
		fail[i + 1] = p;
	}
	int n = strlen(str);
	int s = n / 2 - fail[n / 2];
	int t = fail[n / 2];
	lint l, r; cin >> l >> r;
	auto x = solve(r, s, t);
	auto y = solve(l - 1, s, t);
	for(int i=0; i<26; i++){
		printf("%lld ", x[i] - y[i]);
	}
}
