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<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}

void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define data asd09123jdf02i3h

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++)

const int maxn = 1e6+10;
const int inf = 1e9;
const lf eps = 1e-8;
const int mod = 1e9+7;

struct node {
    int a[26],s;

    node () {memset(a,0,sizeof a);}

    node operator + (const node &x) const {
        node c;
        for(int i=0;i<26;i++) c.a[i]=a[i]+x.a[i];
        c.s=s+x.s;
        return c;
    }

    node operator - (const node &x) const {
        node c;
        for(int i=0;i<26;i++) c.a[i]=a[i]-x.a[i];
        c.s=s-x.s;
        return c;
    }
}f[100];

int tot,n,p,nxt[maxn];
char s[maxn];

void gao() {
    for(int i=2;i<=n;i++) {
        int j=nxt[i-1];
        while(j&&s[j+1]!=s[i]) j=nxt[j];
        if(s[j+1]==s[i]) nxt[i]=j+1;
    }
}

node solve(int r) {
    node ans;
    for(int i=tot;~i;i--) {
        if(r<f[i].s) continue;
        r-=f[i].s,ans=ans+f[i];
    }
    for(int i=1;i<=r;i++) ans.a[s[i]-'a']++;
    return ans;
}

signed main() {
    scanf("%s",s+1),n=strlen(s+1)/2;
    gao();int p=n-nxt[n];
    for(int i=1;i<=n;i++) f[0].a[s[i]-'a']++;f[1]=f[0];
    for(int i=1;i<=p;i++) f[1].a[s[i]-'a']++;
    f[0].s=n,f[1].s=n+p;tot=1;
    for(int i=2;f[i-1].s+f[i-2].s<=1e18;i++) f[++tot]=f[i-1]+f[i-2];
    int l,r;read(l),read(r);
    node ans=solve(r)-solve(l-1);
    for(int i=0;i<26;i++) printf("%lld ",ans.a[i]);puts("");
    return 0;
}