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 <bits/stdc++.h>

using namespace std;

#define ll long long
#define up(i,j,n)	for(int i=j;i<=n;i++)
#define down(i,j,n)	for(int i=j;i>=n;i--)
#define cmax(a,b)	a=max(a,b)
#define cmin(a,b)	a=min(a,b)

const ll MAXN=2e5+5;
const ll oo=2e18+5;
const ll LIM=100;

ll N,l,r,nxt[MAXN],f[105][26],g[105];
char s[MAXN];

namespace solution{
	ll calc(ll p,ll c){
		ll sum=0;
		down(i,LIM,1)if(p>=g[i]){
			sum+=f[i][c];
			p-=g[i];
		}
		up(i,1,p)sum+=((s[i]-'a')==c);
		return sum;
	}
	void Prepare(){
		scanf("%s",s+1);
		scanf("%lld%lld",&l,&r);
		N=strlen(s+1);
		N>>=1;
		ll p=0;
		up(i,2,N){
			while(p&&s[p+1]!=s[i])p=nxt[p];
			if(s[p+1]==s[i])p++;
			nxt[i]=p;
		}
		up(i,1,N)f[1][s[i]-'a']++;
		memcpy(f[2],f[1],sizeof(f[1]));
		up(i,nxt[N]+1,N)f[2][s[i]-'a']++;
		g[1]=N;
		g[2]=N+N-nxt[N];
		up(i,3,LIM){
			g[i]=g[i-1]+g[i-2];
			cmin(g[i],oo);
			up(j,0,25){
				f[i][j]=f[i-1][j]+f[i-2][j];
				cmin(f[i][j],oo);
			}
		}
	}
	void Solve(){
		up(i,0,25)printf("%lld ",calc(r,i)-calc(l-1,i));
		puts("");
	}
}

int main(){
//	freopen("input.in","r",stdin);
	using namespace solution;
	Prepare();
	Solve();
	return 0;
}