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;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

char S[200010];
int n;

int fail[200010];

void Fail(){
	n=strlen(S);
	for(int i=1;i<n/2;++i){
		int j=fail[i-1];
		while(j && S[j]!=S[i]) j=fail[j-1];
		if(S[j]==S[i]) ++j;
		fail[i]=j;
	}
}

ll len[100];
vector<ll> cnt[100];

vector<ll> work(ll to){
	if(!to) return vector<ll>(26);
	vector<ll> ret(26);
	while(to){
		int i=0;
		for(i=0; len[i]; ++i) if(to < len[i]) break;
		if(!i){
			for(int j=0; j<to; ++j) ret[S[j]-'a']++;
			break;
		}
		for(int j=0; j<26; ++j) ret[j]+=cnt[i-1][j];
		to -= len[i-1];
	}
	return ret;
}

int main()
{
	scanf("%s", S);
	Fail();
	len[0] = n/2; cnt[0].resize(26);
	for(int i=0; i<n/2; ++i) cnt[0][S[i]-'a']++;
	cnt[1] = cnt[0]; len[1] = len[0] + n/2-fail[n/2-1];
	for(int i=0; i<n/2-fail[n/2-1]; ++i) cnt[1][S[i]-'a']++;
	
	for(int i=2;;++i){
		len[i] = len[i-1] + len[i-2];
		if(len[i] > ll(1e18)) break;
		cnt[i].resize(26);
		for(int j=0; j<26; ++j) cnt[i][j]=cnt[i-1][j]+cnt[i-2][j];
	}
	
	ll L, R;
	read(L, R);
	auto vl = work(L-1), vr = work(R);
	for(int i=0; i<26; ++i) printf("%lld ", vr[i]-vl[i]);
	return 0;
}