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:

/*\
 * ...
 * ......
 * In the name of ALLAH
 * ......
 * ...
\*/

#include <bits/stdc++.h>

using namespace std;
#define Size(x) ((int)(x).size())
#define pb push_back
#define LD_OUT setprecision(12) << fixed
typedef long long ll;
typedef long double ld;
typedef pair<int,int>pii;
const int INF = 1e9 + 10 , MN = 2e5 + 10 , LG = 222;
const ll BIG = 1000ll*1000*1000*1000*1000*1000 + 10;

int n , nxt[MN];
string s;
ll Nx[LG] , Ny[LG] , X[33][LG] , Y[33][LG];
int ps[33][MN];

void prep()
{
	int cur = 0;
	for(int i=1;i<n;++i){
		while(cur && s[i] != s[cur]) cur = nxt[cur];
		cur += (s[i] == s[cur]);
		nxt[i+1] = cur;
	}
	Nx[0] = n - nxt[n] , Ny[0] = nxt[n];
	for(int i=0;i<Nx[0];++i)
		X[s[i]-'a'][0]++;
	for(int i=Nx[0];i<n;++i)
		Y[s[i]-'a'][0]++;
	for(int i=1;i<LG;++i){
		Nx[i] = Nx[i-1] + Ny[i-1];
		Nx[i] = min(Nx[i] , BIG);
		Ny[i] = Nx[i-1];
		for(int j=0;j<26;++j){
			X[j][i] = X[j][i-1] + Y[j][i-1];
			Y[j][i] = X[j][i-1];
			X[j][i] = min(X[j][i] , BIG);
		}
	}
	for(int i=0;i<n;++i)
		for(int j=0;j<26;++j)
			ps[j][i+1] = ps[j][i] + (s[i]-'a' == j);
}

ll get(ll val,int c)
{
	for(int i=LG-1;~i;--i)
		if(Nx[i] <= val) return min(BIG , X[c][i] + get(val - Nx[i] , c));
	return ps[c][val];
}

int main()
{
	ios_base :: sync_with_stdio(false) ,cin.tie(0) , cout.tie(0);
	cin>>s;
	n = Size(s)/2;
	prep();

	ll l , r;
	cin >> l >> r;
	for(int c=0;c<26;++c) cout << get(r , c) - get(l-1 , c) << ' ';
	cout << '\n';
	return 0;
}