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>
using namespace std;
typedef long long ll;
const int N=200005;
int n,up,nex[N],type;
char str[N];
int pre[N][26];
ll l,r,cut,len[100],ans[26];
ll f[100][26];
void readData(){
    scanf("%s%lld%lld",str+1,&l,&r);
    n=strlen(str+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][str[i]-'a']++;
    }
}
void getNex(){
    nex[1]=0;
    for(int i=2,j;i<=n;i++){
        j=nex[i-1];
        while(j&&str[j+1]!=str[i]) j=nex[j];
        nex[i]=(str[j+1]==str[i])?j+1:0;
    }
}
void preWork(){
    len[1]=n;
    for(int i=1;i<=n;i++){
        f[1][str[i]-'a']++;
        f[2][str[i]-'a']++;
    }
    len[2]=n+cut;
    for(int i=0;i<26;i++) 
        f[2][i]+=pre[cut][i];
    for(int i=3;;i++){
        for(int j=0;j<26;j++) f[i][j]=f[i-2][j]+f[i-1][j];
        len[i]=len[i-2]+len[i-1];
        if(len[i]>1e18){
            up=i;
            break;
        }
    }
}
void calc1(ll m,ll a){
    if(m<=0) return;
    ll t=m/cut;
    for(int i=0;i<26;i++) ans[i]+=a*t*pre[cut][i];
    t=m%cut;        
    for(int i=0;i<26;i++) ans[i]+=a*pre[t][i];
}
void calc2(ll m,ll a){
    for(int i=up;i>=1;i--)
        if(len[i]<=m){
            m-=len[i];
            for(int j=0;j<26;j++) ans[j]+=a*f[i][j];
        }
    for(int j=0;j<26;j++) ans[j]+=a*pre[m][j];
}
int main(){
    readData();
    getNex();
    cut=n-nex[n];
    if(n%cut==0){
        calc1(r,1);
        calc1(l-1,-1);
    }
    else{
        preWork();
        calc2(r,1);
        calc2(l-1,-1);
    }
    for(int i=0;i<26;i++) printf("%lld ",ans[i]);
    return 0;
}