Problem: The Prodiggers are quite a cool band and for this reason, they have been the surprise guest at the ENTER festival for the past 80 years. At the beginning of their careers, they weren’t so successful, so they had to spend time digging channels to earn money; hence the name. Anyway, they like to tour a lot and have surprising amounts of energy to do extremely long tours. However, they hate spending two consecutive days without having a concert, so they would like to avoid it.

A tour is defined by a sequence of concerts and days-off. You need to count in how many ways The Prodiggers can select k different tours of the same length between l and r.

For example if k = 2, l = 1 and r = 2, if we define concert day as {1} and day-off as {0}, here are all possible tours: {0}, {1}, {00}, {01}, {10}, {11}. But tour 00 can not be selected because it has 2 days-off in a row. Now, we need to count in how many ways we can select k = 2 tours of the same length in range [1;2]. Here they are: {0,1}; {01,10}; {01,11}; {10,11}.

Since their schedule is quite busy, they want you to tell them in how many ways can do that, modulo 1 000 000 007 (109 + 7).

Input

The first line of the input contains three integers k, l and r (1 ≤ k ≤ 200, 1 ≤ l ≤ r ≤ 1018).

Output

Output a single number: the number of ways to select k different tours of the same length, modulo 1 000 000 007.

Example

Input

1 1 2


Output

5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const long long mo = (long long)1e9 + 7;
const long long M = 210;
long long S[M][M], C[M][M];
inline long long add(long long p, long long q) {
  return (p += q) >= mo ? p - mo : p;
}
inline long long sub(long long p, long long q) {
  return (p -= q) < 0 ? p + mo : p;
}
inline long long ksm(long long p, long long q, long long mo) {
  long long res = 1;
  for (; q; q >>= 1, p = 1ll * p * p % mo)
    if (q & 1) res = 1ll * res * p % mo;
  return res;
}
inline void init_Stirling() {
  S[0][0] = 1;
  for (long long i = (1); i <= (M - 5); ++i) S[i][0] = 0;
  for (long long i = (1); i <= (M - 5); ++i)
    for (long long j = (1); j <= (i); ++j)
      S[i][j] = sub(S[i - 1][j - 1], 1ll * (i - 1) * S[i - 1][j] % mo);
}
inline void init_C() {
  for (long long i = (0); i <= (M - 5); ++i) C[i][0] = 1;
  for (long long i = (1); i <= (M - 5); ++i)
    for (long long j = (1); j <= (i); ++j)
      C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]);
  cerr << C[7][3] << endl;
}
struct sqt5 {
  long long a, b;
  sqt5() {}
  sqt5(long long _a, long long _b) { a = _a, b = _b; }
  sqt5 operator+(const sqt5 &p) const { return sqt5(add(a, p.a), add(b, p.b)); }
  sqt5 operator-(const sqt5 &p) const { return sqt5(sub(a, p.a), sub(b, p.b)); }
  sqt5 operator*(const sqt5 &p) const {
    return sqt5(add(1ll * a * p.a % mo, 1ll * b * p.b % mo * 5ll % mo),
                add(1ll * b * p.a % mo, 1ll * a * p.b % mo));
  }
  sqt5 operator*(const long long &p) const {
    return sqt5(1ll * a * p % mo, 1ll * b * p % mo);
  }
  sqt5 operator/(const sqt5 &p) const {
    long long iv = ksm(
        sub(1ll * p.a * p.a % mo, 1ll * p.b * p.b % mo * 5ll % mo), mo - 2, mo);
    sqt5 inp = sqt5(1ll * p.a * iv % mo, 1ll * (mo - p.b) * iv % mo);
    return (*this) * inp;
  }
} A, B, invsq5;
inline sqt5 ksm_sqt5(sqt5 p, long long q, long long mo) {
  sqt5 res = sqt5(1, 0);
  for (; q; q >>= 1, p = p * p)
    if (q & 1) res = res * p;
  return res;
}
long long k, l, r, fack;
long long sum(long long n) {
  sqt5 res = sqt5(0, 0);
  for (long long j = (0); j <= (k); ++j) {
    long long eff = S[k][j];
    sqt5 dq = sqt5(0, 0);
    sqt5 ef = ksm_sqt5(invsq5, j, mo);
    for (long long p = (0); p <= (j); ++p) {
      sqt5 q = ksm_sqt5(A, p, mo) * ksm_sqt5(B, j - p, mo);
      sqt5 ret =
          (ksm_sqt5(q, 3, mo) - ksm_sqt5(q, n + 3, mo)) / (sqt5(1, 0) - q);
      if (q.a == 1 && q.b == 0) ret = sqt5(n % mo, 0);
      dq = dq + ret * (1ll * C[j][p] * ((j - p) & 1 ? mo - 1 : 1) % mo);
    }
    res = res + dq * eff * ef;
  }
  return 1ll * res.a * ksm(fack, mo - 2, mo) % mo;
}
int main() {
  init_Stirling();
  init_C();
  A = sqt5((mo + 1) / 2, (mo + 1) / 2),
  B = sqt5((mo + 1) / 2, mo - (mo + 1) / 2);
  invsq5 = sqt5(0, ksm(5, mo - 2, mo));
  k = read();
  l = read(), r = read();
  fack = 1;
  for (long long i = (1); i <= (k); ++i) fack = 1ll * fack * i % mo;
  printf("%lld\n", sub(sum(r), sum(l - 1)));
}