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;
const long long mod = 1000000007;
inline long long cmod(long long x) {
  x %= mod;
  return x < 0 ? x + mod : x;
}
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  if (b & 1) return cmod(bigmod(a, b - 1) * a);
  long long m = bigmod(a, b >> 1);
  return cmod(m * m);
}
inline long long inverse(long long x) {
  x = cmod(x);
  return bigmod(x, mod - 2);
}
class cong {
 public:
  long long a, b;
  cong() {}
  cong(long long a, long long b) {
    this->a = cmod(a);
    this->b = cmod(b);
  }
  cong operator+(const cong &x) const { return cong(a + x.a, b + x.b); }
  cong operator-(const cong &x) const { return cong(a - x.a, b - x.b); }
  cong operator*(const cong &x) const {
    long long c = x.a;
    long long d = x.b;
    return cong((a * c) + (5 * b * d), (a * d) + (b * c));
  }
  cong operator/(long long x) const {
    return cong(a * inverse(x), b * inverse(x));
  }
  cong operator/(const cong g) const {
    long long c = g.a;
    long long d = g.b;
    long long x = cmod(a * c - 5 * b * d) * inverse(c * c - 5 * d * d);
    long long y = cmod(a * d - b * c) * inverse(5 * d * d - c * c);
    return cong(x, y);
  }
  void out() { cout << a << " " << b << endl; }
};
cong bigmod(cong a, long long b) {
  if (b == 0) return cong(1, 0);
  if (b & 1) return bigmod(a, b - 1) * a;
  cong m = bigmod(a, b >> 1);
  return (m * m);
}
cong geosum(cong x, long long n) {
  if (x.a == 1 && x.b == 0) {
    return cong(n + 1, 0);
  }
  return (bigmod(x, n + 1) - cong(1, 0)) / (x - cong(1, 0));
}
cong alpha = cong(1, 1) / 2;
cong beta = cong(1, -1) / 2;
long long c[205][205];
long long a[205][205];
long long fibPower(long long n, int m) {
  cong ans = cong(0, 0);
  for (int j = 0; j <= m; j++) {
    cong t = bigmod(alpha, j) * bigmod(beta, m - j);
    cong x = cong(c[m][j], 0) * geosum(t, n);
    if ((m - j) & 1)
      ans = ans - x;
    else
      ans = ans + x;
  }
  ans = ans * bigmod((cong(0, 1) / 5), m);
  return ans.a;
}
long long query(long long r, int k) {
  if (r < 0) return 0;
  long long ans = 0;
  for (int i = 0; i <= k; i++) {
    long long x = a[k - 1][i] * (fibPower(r + 2, k - i) - 1);
    if (i & 1)
      ans = cmod(ans - x);
    else
      ans = cmod(ans + x);
  }
  for (int i = 1; i <= k; i++) {
    ans *= inverse(i);
    ans %= mod;
  }
  return ans;
}
int main() {
  for (int i = 0; i <= 200; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      c[i][j] %= mod;
    }
  }
  a[0][0] = 1;
  for (int i = 1; i <= 200; i++) {
    a[i][0] = 1;
    for (int j = 1; j <= i + 1; j++) {
      a[i][j] = a[i - 1][j] + i * a[i - 1][j - 1];
      a[i][j] %= mod;
    }
  }
  long long k, l, r;
  cin >> k >> l >> r;
  long long ans = cmod(query(r, k) - query(l - 1, k));
  cout << ans << endl;
}