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>
#include<algorithm>
#define LL long long
using namespace std;
const LL maxn=200010;
const LL inf=1e18;;
char s[maxn];
LL len[1000],nt[1000],cnt1[1000][27],cnt2[1000][27],o[27],ans[27];
LL nxt[maxn];
LL k,t,n,m,tot,top;
LL l,r;
void kmp()
{
    LL j=0;
    for(LL i=2;i<=n;i++)
	{
        while(j&&s[j+1]!=s[i]) j=nxt[j];
        if(s[j+1]==s[i]) j++;
        nxt[i]=j;
    }
}
void work(LL n,LL f)
{
    for(int i=top;i>=1;i--)
        if (n>=len[i])
		{
            for(int j=0;j<26;j++) ans[j]+=f*(cnt1[i][j]*2+cnt2[i][j]);
            n-=len[i];
        }
    if (n>=len[0])
	{
        for(int j=0;j<26;j++) ans[j]+=f*o[j];
        n-=len[0];
    }
    for(int i=1;i<=n;i++) ans[s[i]-'a']+=f;
}
void solve(LL n,LL f)
{
    for(int i=0;i<26;i++) ans[i]+=(LL)f*(n/tot)*o[i];
    n%=tot;
    for(int i=1;i<=n;i++) ans[s[i]-'a']+=f;
}
int main()
{
    scanf("%s",s+1);
    n=strlen(s+1);
    n/=2;
    kmp();
    if (n%(n-nxt[n]))
	{
        for(int i=1;i<=nxt[n];i++) cnt1[0][s[i]-'a']++;
        for(int i=nxt[n]+1;i<=n-nxt[n];i++) cnt2[0][s[i]-'a']++;
        for(int i=1;i<=n;i++) 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];
            for(int i=1;i<=nt[top];i++) cnt1[top][s[i]-'a']++;
            for(int i=n-nxt[n]+1;i<=n;i++) 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];
            for(int i=0;i<26;i++)
			{
                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);
        for(int i=0;i<26;i++) printf("%lld ",ans[i]);
    }
    else
	{
        tot=n-nxt[n];
        for(int i=1;i<=tot;i++) o[s[i]-'a']++;
        scanf("%lld%lld",&l,&r);
        solve(r,1);solve(l-1,-1);
        for(int i=0;i<26;i++) printf("%lld ",ans[i]);
    }
}