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 M = 1e9 + 7;
const int maxk = 225;
int s[maxk], c[maxk][maxk];
inline int nml(long long a) { return ((a % M) + M) % M; }
void exgcd(int a, long long& x, int b, long long& y, int& g) {
  if (b) {
    exgcd(b, y, a % b, x, g);
    y -= a / b * x;
  } else {
    x = 1;
    y = 0;
    g = a;
  }
}
inline int inv(int a) {
  long long x, y;
  int g;
  exgcd(M, x, a, y, g);
  return nml(y);
}
inline int add(int a, int b) { return (a += b) >= M ? a - M : a; }
inline int sub(int a, int b) { return (a -= b) < 0 ? a + M : a; }
inline int mul(int a, int b) { return a * 1LL * b % M; }
inline int _div(int a, int b) { return mul(a, inv(b)); }
inline int mns(int a) { return M - a; }
struct mint {
  int a, b;
  mint(int x, int y) : a(x), b(y) {}
  mint operator+(int x) { return mint(add(a, x), b); }
  mint operator-(int x) { return mint(sub(a, x), b); }
  mint operator*(int x) { return mint(mul(a, x), mul(b, x)); }
  mint operator/(int x) {
    x = inv(x);
    return *this * x;
  }
  mint operator+(mint x) { return mint(add(a, x.a), add(b, x.b)); }
  mint operator-(mint x) { return mint(sub(a, x.a), sub(b, x.b)); }
  mint operator*(mint x) {
    return mint(add(mul(a, x.a), mul(5, mul(b, x.b))),
                add(mul(a, x.b), mul(b, x.a)));
  }
  mint operator/(mint x) {
    int v = sub(mul(x.a, x.a), mul(5, mul(x.b, x.b)));
    return *this * mint(x.a, mns(x.b)) / v;
  }
  mint operator*=(mint x) { return *this = *this * x; }
  mint operator+=(mint x) { return *this = *this + x; }
  mint operator/=(int x) { return *this = *this / x; }
  bool operator!=(int x) { return b || a != x; }
};
mint operator*(int x, mint y) { return y * x; }
mint operator/(int x, mint y) { return mint(x, 0) / y; }
void precalc(int k) {
  s[1] = 1;
  for (int i = 2; i <= k; ++i)
    for (int j = i; j >= 1; --j) s[j] = add(mul(mns(i - 1), s[j]), s[j - 1]);
  for (int i = 1; i <= k; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; ++j) c[i][j] = add(c[i - 1][j], c[i - 1][j - 1]);
  }
}
mint qpow(mint a, long long x) {
  int sig = x < 0 ? -1 : 1;
  x *= sig;
  mint ret(1, 0);
  for (; x; x >>= 1, a *= a)
    if (x & 1) ret *= a;
  return (sig > 0) ? ret : (1 / ret);
}
int main() {
  int k;
  long long l, r;
  scanf("%d%lld%lld", &k, &l, &r);
  {
    int a = 0, b = 1, i = 1;
    for (; b < k; ++i, b += a, a = b - a)
      ;
    if (l + 2 < i) l = i - 2;
    if (r < l) {
      puts("0");
      return 0;
    }
  }
  precalc(k);
  mint isr5 = 1 / mint(0, 1);
  mint p = mint(1, 1) / 2;
  mint ip = 1 / p;
  mint p2 = p * p;
  mint pr3 = qpow(p, r + 3);
  mint pl2 = qpow(p, l + 2);
  mint ipr3 = 1 / pr3;
  mint ipl2 = 1 / pl2;
  mint pr3_2 = pr3 * pr3;
  mint pl2_2 = pl2 * pl2;
  mint rng(nml(r - l + 1), 0);
  mint ans(0, 0);
  mint v1 = isr5;
  mint v2 = ip;
  mint v3 = ipr3;
  mint v4 = ipl2;
  for (int j = 1; j <= k; ++j, v1 *= isr5, v2 *= ip, v3 *= ipr3, v4 *= ipl2) {
    mint sum(0, 0);
    int ne = (j & 1) ? mns(1) : 1;
    mint vv2 = v2;
    mint vv3 = v3;
    mint vv4 = v4;
    for (int m = 0; m <= j;
         ++m, ne = mul(ne, mns(1)), vv2 *= p2, vv3 *= pr3_2, vv4 *= pl2_2) {
      mint q = ne * vv2;
      mint qr3 = (((r + 3) & 1) ? ne : 1) * vv3;
      mint ql2 = (((l + 2) & 1) ? ne : 1) * vv4;
      sum += mul(ne, c[j][m]) * (q != 1 ? (qr3 - ql2) / (q - 1) : rng);
    }
    ans += s[j] * v1 * sum;
  }
  int ik = 1;
  for (int i = 1; i <= k; ++i) ik = mul(ik, i);
  printf("%d\n", _div(ans.a, ik));
}