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
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cctype>
#include<cstdlib>
#include<cstring>
#include<utility>
#include<algorithm>
 
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define For(i,x,y) for(int i=((int)x);i<=((int)y);i++)
#define Dep(i,y,x) for(int i=((int)y);i>=((int)x);i--)
#define Rep(i,x) for (int y,i=head[x];i;i=E[i].nxt)
using namespace std;
 
typedef long long ll;
typedef unsigned int uint;
typedef pair<int,int> pii;
 
int rd() {
    char c=getchar(); int t=0,f=1;
    while (!isdigit(c)) f=(c=='-')?-1:1,c=getchar();
    while (isdigit(c)) t=t*10+c-48,c=getchar(); return t*f;
}
void wt(int x) {
    if (x<0) putchar('-'),wt(-x);
    else { if (x>9) wt(x/10); putchar(x%10+48); }
}

const int N=2e5+5;

char s[N];
int n,j,t,up,fail[N];
ll l,r,ans[26],len[101],pre[N][26],sum[101][26];

void solve(ll p,ll f) {
	for (int i=up;i>=0;i--) if (p>=len[i]) {
		p-=len[i];
		For (j,0,25) ans[j]+=f*sum[i][j];
	}
	For (j,0,25) ans[j]+=f*pre[p][j];
}

int main() {
	scanf("%s",s+1);
	scanf("%lld%lld",&l,&r);
	
	n=strlen(s+1)/2,j=0;
	For (i,2,n) {
		while (j && s[j+1]!=s[i]) j=fail[j];
		if (s[j+1]==s[i]) j++; fail[i]=j;
	}
	t=n-fail[n];
	For (i,1,n) For (j,0,25)
		pre[i][j]=pre[i-1][j]+(s[i]=='a'+j);
	
	len[0]=n;
	len[1]=n+t;
	For (j,0,25) {
		sum[0][j]=pre[n][j];
		sum[1][j]=pre[n][j]+pre[t][j];
	}
	for (up=2;len[up-2]+len[up-1]<=1e18;up++) {
		len[up]=len[up-2]+len[up-1];
		For (j,0,25) sum[up][j]=sum[up-2][j]+sum[up-1][j];
	}
	up--;
	
	solve(l-1,-1);
	solve(r,1);
	
	For (j,0,25) printf("%lld ",ans[j]); puts("");
	return 0;
}