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:
/*Program from Luvwgyx*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll  inf=1e18;
const int maxn=2e5+10;
const int maxm=1e2+10;
char s[maxn];int n,lst,nxt[maxn],pre[maxn][26];
ll l,r,cutpos,ans[26],cnt[maxm],f[maxm][26];
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void print(ll x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)print(x/10);
    putchar(x%10+'0');
}
void write(ll x){print(x);putchar(' ');}
void get_nxt(){
    nxt[1]=0;
    for(int i=2,j;i<=n;i++){
        j=nxt[i-1];
        while(j&&s[j+1]!=s[i])j=nxt[j];
        nxt[i]=s[j+1]==s[i]?j+1:0;
    }
}
void cal1(ll x,ll v){
    if(x<=0)return ;ll tmp=x/cutpos;
    for(int i=0;i<26;i++)ans[i]+=v*tmp*pre[cutpos][i];
    tmp=x%cutpos;
    for(int i=0;i<26;i++)ans[i]+=v*pre[tmp][i];
}
void get_fib(){
    cnt[1]=n;
    for(int i=1;i<=n;i++){f[1][s[i]-'a']++;f[2][s[i]-'a']++;}
    cnt[2]=n+cutpos;
    for(int i=0;i<26;i++)f[2][i]+=pre[cutpos][i];
    for(int i=3;;i++){
        for(int j=0;j<26;j++)f[i][j]=f[i-2][j]+f[i-1][j];
        cnt[i]=cnt[i-2]+cnt[i-1];if(cnt[i]>inf){lst=i;break;}
    }
}
void cal2(ll x,ll v){
    for(int i=lst;i>=1;i--)
        if(cnt[i]<=x){
            x-=cnt[i];
            for(int j=0;j<26;j++)ans[j]+=v*f[i][j];
        }
    for(int i=0;i<26;i++)ans[i]+=v*pre[x][i];
}
int main(){
    scanf("%s",s+1);l=read();r=read();n=strlen(s+1);n>>=1;
    for(int i=1;i<=n;i++){
        for(int j=0;j<26;j++)pre[i][j]+=pre[i-1][j];
        pre[i][s[i]-'a']++;
    }get_nxt();cutpos=n-nxt[n];//write(cutpos);puts("");
    //if(!n%cutpos){cal1(r,1);cal1(l-1,-1);}
    //else {get_fib();cal2(r,1);cal2(l-1,-1);}
    get_fib();cal2(r,1);cal2(l-1,-1);
    for(int i=0;i<26;i++)write(ans[i]);puts("");
    return 0;
}