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;
#define rep(i,a,b) for(int i = a; i < (b); ++i)
#define rrep(i,a,b) for(int i = b; i --> (a);)
#define all(v) v.begin(),v.end()
#define trav(x,v) for(auto &x : v)
#define sz(v) (int)(v).size()
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

const ll md = 1e9+7;

const ll b = 2;//1342231LL;

const int mx = 2e5;

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	vector<ll> pw(mx);
	pw[0] = 1;
	rep(i,1,mx) pw[i] = pw[i-1]*b % md;

	string s;
	cin >> s;
	int n = sz(s)/2;
	string a = s.substr(0, n);

	vector<ll> pre(n+1);
	rep(i,0,n) pre[i+1] = (pre[i]*b + a[i])%md;

	int len;
	for(len = n; len --> 0;){
		if((pre[n]-pre[n-len]*pw[len]%md - pre[len])%md == 0)
			break;
	}
	ll l, r;
	cin >> l >> r;

	vector<ll> fibs(200);
	fibs[0] = len, fibs[1] = n-len;
	rep(i,2,200) fibs[i] = fibs[i-1] + fibs[i-2];

	vector<vector<ll>> cnts(200, vector<ll>(26));
	rep(i,0,26) cnts[0][i] = count(a.begin(),a.begin()+len, 'a'+i);
	rep(i,0,26) cnts[1][i] = count(a.begin(),a.begin()+n-len, 'a'+i);
	rep(i,2,200) rep(j,0,26) cnts[i][j] = cnts[i-1][j]+cnts[i-2][j];

	function<ll(ll,int,int)> que = [&](ll pos, int c, int ix){
		assert(fibs[ix] >= pos);
		if(ix <= 1){
			return (ll)count(a.begin(),a.begin()+pos, 'a'+c);
		}
		if(pos <= fibs[ix-1]) return que(pos, c, ix-1);
		return cnts[ix-1][c] + que(pos-fibs[ix-1],c,ix-2);
	};

	function<ll(ll,int)> f = [&](ll pos, int c){
		int i = 0;
		while(fibs[i] < pos) ++i;
		return que(pos, c, i);
	};
	rep(c,0,26){
		cout << f(r,c) - f(l-1,c) << ' ';
	}
	cout << endl;
}