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;

typedef long long ll;

void getPi(string s, vector<int> &pi) {
    pi = vector<int>(s.size(), 0);
    int j = 0;
    for(int i = 1; i < s.size(); i++) {
        while(j && s[i] != s[j]) j = pi[j - 1];
        if(s[i] == s[j]) pi[i] = ++j;
    }
}

string S, T, A, B;
ll L, R;
vector<int> pi;
vector<ll> fibo;
vector<vector<ll> > fcnt;

void Fibonacci() {
    //cout << A << ' ' << B << endl;
    fibo.push_back(A.size());
    fibo.push_back(B.size());
    for(int i = 2;; i++) {
        ll a = fibo[i - 2];
        ll b = fibo[i - 1];
        fibo.push_back(a + b);
        if(fibo[i] >= 1e18) break;
    }
    fcnt = vector<vector<ll> >(fibo.size(), vector<ll>(26, 0));
    for(int i = 0; i < A.size(); i++) {
        fcnt[0][ A[i] - 'a' ]++;
    }
    for(int i = 0; i < B.size(); i++) {
        fcnt[1][ B[i] - 'a' ]++;
    }
    for(int i = 2; i < fibo.size(); i++) {
        for(int j = 0; j < 26; j++) {
            fcnt[i][j] = fcnt[i - 1][j] + fcnt[i - 2][j];
        }
    }
}

vector<ll> solve(ll x) {
    if(x < 0) return vector<ll>(26, 0);
    if(x + 1 <= B.size()) {
        vector<ll> ret(26, 0);
        for(int i = 0; i <= x; i++) {
            ret[ B[i] - 'a' ]++;
        }
        return ret;
    }

    int s = 0, e = fibo.size() - 1;
    int r;
    while(s <= e) {
        int m = (s + e)>>1;

        if(fibo[m] <= x + 1) {
            r = m;
            s = m + 1;
        }
        else e = m - 1;
    }
    vector<ll> ret(26, 0);
    for(int i = 0; i < 26; i++) {
        ret[i] += fcnt[r][i];
    }

    vector<ll> tmp = solve(x - fibo[r]);
    for(int i = 0; i < 26; i++) {
        ret[i] += tmp[i];
    }
    return ret;
}

int main() {
    cin>>S>>L>>R;
    L--; R--;

    for(int i = 0; i < S.size() / 2; i++) T.push_back(S[i]);
    getPi(T, pi);

    A = T;
    for(int i = 0; i < (int)T.size() - pi[T.size() - 1]; i++) A.push_back(T[i]);
    B = A;
    for(int i = 0; i < T.size(); i++) B.push_back(T[i]);

    //cout << pi[T.size() - 1] << ' ' << T.size() << endl;

    //cout << A << ' ' << B << endl;

    Fibonacci();

    vector<ll> Y = solve(R);
    vector<ll> X = solve(L - 1);

    for(int i = 0; i < 26; i++) {
        printf("%lld ", Y[i] - X[i]);
    }
}