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 <cstdio>
#include <cstring>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)
using namespace std;

typedef long long LL;
int n,m,nxt[200100];
LL l,r,ans[26],sum[100][26],len[100];
char a[200100];

int calc(int m)
{
    int x=m-n%m;
    return (n+x)/m&1?x+m:x;
}

void solve(LL n,int w)
{
    repd(i,m,0)
        if (n>=len[i])
        {
            n-=len[i];
            repu(j,0,25)
                ans[j]+=w*sum[i][j];
        }
    repu(i,1,n)
        ans[a[i]-'a']+=w;
}

int main()
{
    scanf("%s%lld%lld",a+1,&l,&r),n=strlen(a+1)/2;
    for (int i=2,j=0; i<=n; ++i)
    {
        for (; j && a[i]!=a[j+1]; j=nxt[j]);
        nxt[i]=j+=a[i]==a[j+1];
    }
    len[m=1]=n,len[0]=n-nxt[n];
    repu(i,1,n)
    {
        ++sum[1][a[i]-'a'];
        sum[0][a[i]-'a']+=i<=len[0];
    }
    while (len[m]<r)
    {
        ++m,len[m]=len[m-1]+len[m-2];
        repu(i,0,25)
            sum[m][i]=sum[m-1][i]+sum[m-2][i];
    }
    solve(l-1,-1),solve(r,1);
    repu(i,0,25)
        printf("%lld ",ans[i]);
    return 0;
}