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 double db;
typedef long long ll;
#define w1 first
#define w2 second
#define ls (x<<1)
#define rs (x<<1|1)
#define pb push_back
#define mid ((l+r)>>1)
#define SZ(x) ((x).size())
#define All(x) (x).begin(),(x).end()
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
	num=0;T f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
	num*=f;
}
const int maxn=2e5+5;
int totlen;
ll l,r;
ll len[maxn];
int nxt[maxn];
char str[maxn];
struct val{
	ll cnt[26];
	val(){memset(cnt,0,sizeof cnt);}
	friend val operator+(val A,val B){
		val C;
		rep2(i,0,26)C.cnt[i]=A.cnt[i]+B.cnt[i];
		return C;
	}
	friend val operator-(val A,val B){
		val C;
		rep2(i,0,26)C.cnt[i]=A.cnt[i]-B.cnt[i];
		return C;
	}
	friend void Print(val A){
		rep2(i,0,26)printf("%lld ",A.cnt[i]);puts("");
	}
}v[105];
void getnxt(){
	int tmp=0;
	rep(i,2,totlen){
		while(tmp&&str[tmp+1]!=str[i])tmp=nxt[tmp];
		if(str[tmp+1]==str[i])tmp++;
		nxt[i]=tmp;
	}
	for(int i=nxt[totlen];;i=nxt[i])if(i<(totlen+1)/2){
		len[1]=totlen-i;
		break;
	}
	len[0]=len[1]-nxt[len[1]];
}
val Calc(int now,ll pos){
	val res;
	if(now==1){
		rep(i,1,pos)res.cnt[str[i]-'a']++;
		return res;
	}
	if(pos<=len[now-1])return Calc(now-1,pos);
	return v[now-1]+Calc(now-1,pos-len[now-1]);
}
int main(){
	scanf("%s",str+1);
	totlen=strlen(str+1);
	getnxt();
	rep(i,1,len[0])v[0].cnt[str[i]-'a']++;
	rep(i,1,len[1])v[1].cnt[str[i]-'a']++;
	read(l);read(r);
	int maxi=1;
	rep(i,2,100){
		len[i]=len[i-1]+len[i-2],v[i]=v[i-1]+v[i-2];
		if(len[i]>r){maxi=i;break;}
	}
	Print(Calc(maxi,r)-Calc(maxi,l-1));
	return 0;
}