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 <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <sstream>
#include <complex>
#include <ctime>
#include <cassert>
#include <functional>

using namespace std;

typedef long long ll;
typedef vector<int> VI;
typedef pair<int,int> PII;

#define REP(i,s,t) for(int i=(s);i<(t);i++)
#define FILL(x,v) memset(x,v,sizeof(x))

const ll INF = (ll)1E18+1;
#define MAXN 200005

char pattern[MAXN];
int nxt[MAXN];
void calcNext(){
  int k=0, len = strlen(pattern);
  nxt[0] = 0;
  for(int i=1; i<=len; i++){
    if(pattern[i]==pattern[k]) nxt[i] = nxt[k++];
    else{
      nxt[i] = k;
      while(k>0 && pattern[i]!=pattern[k]) k=nxt[k];
      if(pattern[k]==pattern[i]) k++;
    }
  }
}

ll plen[100], cnt[100][26];
int N, Tlen, stcnt[MAXN][26];
vector<ll> solve(int i, ll k) {
  vector<ll> sol(26,0), sol2(26,0);
  if (k == 0) return sol;
  if (i <= 1) {
    REP(c,0,26) sol[c] = stcnt[k-1][c];
    return sol;
  }
  if (k <= plen[i-1]) {
    sol = solve(i-1, k);
  } else {
    REP(c,0,26) sol[c] = cnt[i-1][c];
    sol2 = solve(i-2, k-plen[i-1]);
    REP(c,0,26) sol[c] += sol2[c];
  }
  return sol;
}
int main() {
  ll L, R;
  cin >> pattern >> L >> R;
  N = strlen(pattern);
  N /= 2;
  pattern[N] = 0;
  calcNext();
  Tlen = N - nxt[N];
  REP(i,N,N+Tlen) pattern[i] = pattern[i-N];
  REP(i,0,N+Tlen) {
    if (i) REP(c,0,26) stcnt[i][c] = stcnt[i-1][c];
    stcnt[i][pattern[i]-'a']++;
  }
  plen[0] = N;
  plen[1] = N + Tlen;
  REP(c,0,26) {
    cnt[0][c] = stcnt[plen[0]-1][c];
    cnt[1][c] = stcnt[plen[1]-1][c];
  }
  int maxi = 1;
  REP(i,2,100) {
    plen[i] = plen[i-1] + plen[i-2];
    REP(c,0,26) cnt[i][c] = cnt[i-1][c] + cnt[i-2][c];
    if (plen[i] >= INF) {
      maxi = i;
      break;
    }
  }
  vector<ll> ans = solve(maxi, R);
  vector<ll> dec = solve(maxi, L-1);
  REP(c,0,26) cout << ans[c] - dec[c] << " ";
  return 0;
}