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 int maxn = 222, mod = 1000000007, inv2 = 500000004, inv5 = 400000003;
inline unsigned long long read() {
  char ch = getchar();
  unsigned long long x = 0, f = 0;
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int k, fac[maxn], invfac[maxn], ans, S[maxn][maxn], C[maxn][maxn];
unsigned long long l, r;
inline int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
inline int sub(int x, int y) { return x < y ? x - y + mod : x - y; }
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline int qpow(int a, unsigned long long b) {
  int ans = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ans = mul(ans, a);
  return ans;
}
struct comp {
  int x, y;
  comp(const int xx = 0, const int yy = 0) : x(xx), y(yy) {}
  inline comp operator+(const comp &c) const {
    return comp(add(x, c.x), add(y, c.y));
  }
  inline comp operator-(const comp &c) const {
    return comp(sub(x, c.x), sub(y, c.y));
  }
  inline comp operator*(const comp &c) const {
    return comp(add(mul(x, c.x), mul(5, mul(y, c.y))),
                add(mul(x, c.y), mul(y, c.x)));
  }
  inline comp inv() const {
    comp ans(x, y ? mod - y : 0);
    int dn = qpow(sub(mul(x, x), mul(5, mul(y, y))), mod - 2);
    return ans * dn;
  }
  inline comp operator/(const comp &c) const { return *this * c.inv(); }
  inline bool operator==(const comp &c) const { return x == c.x && y == c.y; }
} a(0, inv5), b(0, mod - inv5), x(inv2, inv2), y(inv2, mod - inv2);
inline comp cqpow(comp a, unsigned long long b) {
  comp ans(1, 0);
  for (; b; b >>= 1, a = a * a)
    if (b & 1) ans = ans * a;
  return ans;
}
comp calc(comp x, unsigned long long l, unsigned long long r) {
  if (x == 1) return (r - l + 1) % mod;
  return (cqpow(x, r + 1) - cqpow(x, l)) / (x - 1);
}
int main() {
  k = read();
  l = read();
  r = read();
  for (int i = (0); i <= (k); i++) C[i][0] = C[i][i] = 1;
  for (int i = (1); i <= (k); i++)
    for (int j = (1); j <= (i - 1); j++)
      C[i][j] = add(C[i - 1][j], C[i - 1][j - 1]);
  S[1][1] = 1;
  for (int i = (2); i <= (k); i++)
    for (int j = (1); j <= (i); j++)
      S[i][j] = add(mul(i - 1, S[i - 1][j]), S[i - 1][j - 1]);
  fac[0] = 1;
  for (int i = (1); i <= (k); i++) fac[i] = mul(fac[i - 1], i);
  invfac[k] = qpow(fac[k], mod - 2);
  for (int i = (0); i <= (k); i++) {
    int s = 0;
    for (int j = (0); j <= (i); j++) {
      comp tmp1 = cqpow(a, j) * cqpow(b, i - j),
           tmp2 = cqpow(x, j) * cqpow(y, i - j);
      s = add(s, mul(C[i][j], (tmp1 * calc(tmp2, l + 2, r + 2)).x));
    }
    s = mul(s, S[k][i]);
    if ((k - i) & 1)
      ans = sub(ans, s);
    else
      ans = add(ans, s);
  }
  printf("%d\n", mul(ans, invfac[k]));
}