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<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}


vector<int>Z_Algorithm(const string&s) {
    int n=s.size();
    vector<int>z(n);
    int l = 0, r = 0;
    for (int i=1; i<n; i++) {
	if (i > r) {
	    l = r = i;
	    while (r<n && s[r-l] == s[r]) r++;
	    z[i] = r-l; r--;
	} else {
	    int k = i-l;
	    if (z[k] < r-i+1) z[i] = z[k];
	    else {
		l = i;
		while (r<n && s[r-l] == s[r]) r++;
		z[i] = r-l; r--;
	    }
	}
    }
    return z;
}
const int LEVEL = 100;
const LL INF = 1LL<<61;
int N;
char S[200011];
LL C[LEVEL][26];
LL len[LEVEL];
LL ans[26];

void F(LL R, int sgn) {
    for (int i=LEVEL-1; i>=2; i--) {
	if (len[i] <= R) {
	    REP (t, 26) ans[t] += C[i][t] * sgn;
	    R -= len[i];
	}
    }

    assert(R <= N);
    REP (i, R) ans[S[i]-'a'] += sgn;
}

void MAIN() {
    scanf("%s", S);
    N = strlen(S);
    N /= 2;
    VI z = Z_Algorithm(string(S, S+N));
    
    int pos = N;
    for (int i=1; i<N; i++) if (i + z[i] == N) {
	pos = i;
	break;
    }

    len[0] = pos;
    len[1] = N;
    REP (i, N) {
	if (i < pos) C[0][S[i]-'a']++;
	C[1][S[i]-'a']++;
    }
    for (int i=2; i<LEVEL; i++) {
	len[i] = min(INF, len[i-1] + len[i-2]);
	REP (t, 26) C[i][t] = C[i-1][t] + C[i-2][t];
    }
//    puts(S);

    LL l, r;
    scanf("%lld%lld", &l, &r);
    assert(len[LEVEL-1] > r);
    F(r, 1);
    F(l-1, -1);

    //puts(S);
    //printf("%d %d\n", sep, pos);
    rprintf("%lld", ans, ans+26);
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}