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<cstdio>
#include<algorithm>
#include<cstring>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int maxn=200000+10;
const ll inf=1000000000000000000;
char s[maxn];
ll len[1000],nt[1000],cnt1[1000][27],cnt2[1000][27],o[27],ans[27];
int nxt[maxn];
int i,j,k,t,n,m,tot,top;
ll l,r;
void kmp(){
	j=0;
	fo(i,2,n){
		while (j&&s[j+1]!=s[i]) j=nxt[j];
		if (s[j+1]==s[i]) j++;
		nxt[i]=j;
	}
}
void work(ll n,int f){
	int i,j;
	fd(i,top,1)
		if (n>=len[i]){
			fo(j,0,25) ans[j]+=f*(cnt1[i][j]*2+cnt2[i][j]);
			n-=len[i];
		}
	if (n>=len[0]){
		fo(j,0,25) ans[j]+=f*o[j];
		n-=len[0];
	}
	fo(i,1,n) ans[s[i]-'a']+=f;
}
void solve(ll n,int f){
	int i,j;
	fo(i,0,25) ans[i]+=(ll)f*(n/tot)*o[i];
	n%=tot;
	fo(i,1,n) ans[s[i]-'a']+=f;
}
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	n/=2;
	kmp();
	if (n%(n-nxt[n])){
		fo(i,1,nxt[n]) cnt1[0][s[i]-'a']++;
		fo(i,nxt[n]+1,n-nxt[n]) cnt2[0][s[i]-'a']++;
		fo(i,1,n) o[s[i]-'a']++;
		len[0]=n;nt[0]=nxt[n];
		if (nxt[n]>=(n+1)/2){
			top++;
			len[top]=len[top-1]*2-nt[top-1];
			nt[top]=len[top-1]-nt[top-1];
			fo(i,1,nt[top]) cnt1[top][s[i]-'a']++;
			fo(i,n-nxt[n]+1,n) cnt2[top][s[i]-'a']++;
		}
		while (len[top]<inf){
			top++;
			len[top]=len[top-1]*2-nt[top-1];
			nt[top]=len[top-1]-nt[top-1];
			fo(i,0,25){
				cnt1[top][i]=cnt1[top-1][i]+cnt2[top-1][i];
				cnt2[top][i]=cnt1[top-1][i];
			}
		}
		scanf("%lld%lld",&l,&r);
		work(r,1);
		work(l-1,-1);
		fo(i,0,25) printf("%lld ",ans[i]);
	}
	else{
		tot=n-nxt[n];
		fo(i,1,tot) o[s[i]-'a']++;
		scanf("%lld%lld",&l,&r);
		solve(r,1);
		solve(l-1,-1);
		fo(i,0,25) printf("%lld ",ans[i]);
	}
}