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 <cstring>
typedef long long ll;

int n,nk,f[100005],g[100005][26];
char s[200005];

ll Calc(ll m,int c)
{
    if(m<=n)return g[m][c];
    ll l1=n,l2=nk,k,s1=g[l1][c],s2=g[l2][c];
    while(l1+l2<=m)k=l1,l1+=l2,l2=k,k=s1,s1+=s2,s2=k;
    return s1+Calc(m-l1,c);
}

int main()
{
    scanf("%s",s+1),n=strlen(s+1)>>1,f[0]=-1;
    for(int i=1,j=-1;i<=n;f[i++]=++j)while(j>=0&&s[j+1]!=s[i])j=f[j];
    nk=n-f[n];
    for(int i=1;i<=n;++i)memcpy(g[i],g[i-1],sizeof g[i]),++g[i][s[i]-'a'];
    ll l,r;
    scanf("%lld%lld",&l,&r);
    for(int i=0;i<26;++i)printf("%lld%c",Calc(r,i)-Calc(l-1,i),i==25?'\n':' ');
    return 0;
}