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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define forE(i,x) for(int i=head[x];i!=-1;i=ne[i])
using namespace std;
typedef long long i64;
typedef unsigned long long u64;
typedef unsigned u32;
typedef pair<int,int> pin;
#define mk(a,b) make_pair(a,b)
#define lowbit(x) ((x)&(-(x)))
#define sqr(a) ((a)*(a))
#define clr(a) (memset((a),0,sizeof(a)))
#define ls ((x)<<1)
#define rs (((x)<<1)|1)
#define mid (((l)+(r))>>1)
#define pb push_back
#define w1 first
#define w2 second
inline void read(int &x){
	x=0;int f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	x*=f;
}
inline void judge(){
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
}
/*******************************head*******************************/
const int maxn=300005;
int nxt[maxn],n,k;
char s[maxn];
inline i64 calc(int c,i64 r){
	i64 res=0;
	if(r<=n){
		rep(i,1,r)res+=s[i]==c;
		return res;
	}
	i64 ca=0,cb=0;
	rep(i,1,n-k)ca+=s[i]==c;
	rep(i,n-k+1,n)cb+=s[i]==c;
	if(k%(n-k)==0){
		i64 cc=0;
		rep(i,n-k+1,n-k+(r-(n-k))%k)cc+=s[i]==c;
		return ca+cb*((r-(n-k))/k)+cc;
	}else{
		i64 a=n-k,b=k;
		for(;a+b<=r;b+=a,swap(a,b),cb+=ca,swap(ca,cb));
		return ca+calc(c,r-a);
	}
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1)/2;
	rep(i,2,n){
		int p=nxt[i-1];
		while(p&&s[p+1]!=s[i])p=nxt[p];
		if(s[p+1]==s[i])p++;
		nxt[i]=p;
	}
	k=nxt[n];
	if(k==0)k=n,n+=n;
	i64 l,r;
	cin>>l>>r;
	rep(i,0,25)cout<<calc(i+'a',r)-calc(i+'a',l-1)<<' ';
	return 0;
}