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>
#pragma comment(linker, "/STACK:128000000")
using namespace std;
void solve();
void precalc();
int TESTNUM = 0;
int main() {
  string s = "";
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  precalc();
  int t = 1;
  while (t--) {
    ++TESTNUM;
    solve();
  }
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
void precalc() {}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
inline int nxt() {
  int x;
  cin >> x;
  return x;
}
const long long mod = 1000000007;
struct Num {
  long long a, b;
  Num(long long x = 0, long long y = 0) {
    a = x;
    b = y;
  }
  Num operator+(const Num& other) const {
    long long na = a + other.a;
    long long nb = b + other.b;
    if (na >= mod) na -= mod;
    if (nb >= mod) nb -= mod;
    return Num(na, nb);
  }
  Num operator-(const Num& other) const {
    long long na = a - other.a;
    long long nb = b - other.b;
    if (na < 0) na += mod;
    if (nb < 0) nb += mod;
    return Num(na, nb);
  }
  Num operator*(const Num& other) const {
    long long na = (1ll * a * other.a + 5ll * b * other.b) % mod;
    long long nb = (1ll * a * other.b + 1ll * b * other.a) % mod;
    if (na < 0) na += mod;
    if (nb < 0) nb += mod;
    return Num(na, nb);
  }
  long long norm() const {
    long long tmp = (1ll * a * a - 5ll * b * b) % mod;
    if (tmp < 0) tmp += mod;
    return tmp;
  }
  Num overline() const { return (b == 0) ? *this : Num(a, mod - b); }
  Num operator/(const Num& other) const {
    long long inv = other.norm();
    inv = binpow<long long>(inv, mod - 2, mod);
    Num res = (*this * other.overline());
    res.a = 1ll * res.a * inv % mod;
    res.b = 1ll * res.b * inv % mod;
    return res;
  }
  void out() const {
    if (a && b) {
      if (b == 1)
        cerr << a << " + sqrt(5)\n";
      else
        cerr << a << " + " << b << " * sqrt(5)\n";
    } else {
      if (b) {
        if (b == 1)
          cerr << "sqrt(5)\n";
        else
          cerr << b << " * sqrt(5)\n";
      } else
        cerr << a << "\n";
    }
  }
};
Num pw(Num a, long long n) {
  Num res(1);
  while (n) {
    if (n & 1ll) res = res * a;
    n >>= 1;
    a = a * a;
  }
  return res;
}
Num sum(const Num& x, long long l, long long r) {
  if (x.a == 1 && x.b == 0) return (r - l + 1) % mod;
  return pw(x, l) * (pw(x, r - l + 1) - 1) / (x - 1);
}
const long long K = 222;
long long C[K][K];
vector<long long> polymult(const vector<long long>& p1,
                           const vector<long long>& p2) {
  vector<long long> res;
  for (long long i = 0; i < (long long)p1.size(); i++) {
    for (long long j = 0; j < (long long)p2.size(); j++) {
      if (res.size() <= i + j) res.push_back(0);
      res[i + j] += p1[i] * p2[j] % mod;
    }
  }
  for (auto& x : res) {
    x %= mod;
    if (x < 0) x += mod;
  }
  return res;
}
long long stupid(long long k, long long l, long long r) {
  long long index = 1;
  l += 2, r += 2;
  long long x = 0;
  long long y = 1;
  long long ans = 0;
  while (index <= r) {
    if (index >= l) {
      long long cur = 1;
      for (long long i = 0; i < k; i++) {
        cur =
            cur * (y - i) % mod * binpow<long long>(i + 1, mod - 2, mod) % mod;
      }
      ans += cur;
    }
    x = (x + y) % mod;
    swap(x, y);
    index++;
  }
  return ans % mod;
}
long long not_stupid(long long k, long long l, long long r) {
  Num phi_1(1, 1);
  Num phi_2(1, mod - 1);
  phi_1 = phi_1 / 2;
  phi_2 = phi_2 / 2;
  Num denom = Num(1) / Num(0, 1);
  l += 2;
  r += 2;
  for (long long i = 0; i < K; i++) {
    C[i][0] = C[i][i] = 1;
    for (long long j = 1; j < i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      if (C[i][j] >= mod) C[i][j] -= mod;
    }
  }
  vector<Num> sm(k + 1);
  Num curdenom(1);
  for (long long d = 0; d <= k; d++) {
    Num cur = pw(phi_1, d);
    long long co = 1;
    for (long long i = 0; i <= d; i++) {
      sm[d] = sm[d] + sum(cur, l, r) * C[d][i] * co;
      co = mod - co;
      cur = cur * phi_2 / phi_1;
    }
    sm[d] = sm[d] * curdenom;
    curdenom = curdenom * denom;
  }
  vector<long long> poly(1, 1);
  for (long long i = 0; i < k; i++) {
    poly = polymult(poly, {-i, 1});
  }
  Num ans(0);
  for (long long i = 0; i <= k; i++) {
    ans = ans + sm[i] * poly[i];
  }
  long long fact = 1;
  for (long long i = 1; i <= k; i++) fact = fact * i % mod;
  return (ans / fact).a;
}
void solve() {
  long long k;
  cin >> k;
  long long l, r;
  cin >> l >> r;
  long long res = not_stupid(k, l, r) % mod;
  if (res < 0) {
    res += mod;
  }
  cout << res << "\n";
}