Problem: For given n, l and r find the number of distinct geometrical progression, each of which contains n distinct integers not less than l and not greater than r. In other words, for each progression the following must hold: l ≤ ai ≤ r and ai ≠ aj , where a1, a2, ..., an is the geometrical progression, 1 ≤ i, j ≤ n and i ≠ j.

Geometrical progression is a sequence of numbers a1, a2, ..., an where each term after first is found by multiplying the previous one by a fixed non-zero number d called the common ratio. Note that in our task d may be non-integer. For example in progression 4, 6, 9, common ratio is <image>.

Two progressions a1, a2, ..., an and b1, b2, ..., bn are considered different, if there is such i (1 ≤ i ≤ n) that ai ≠ bi.

Input

The first and the only line cotains three integers n, l and r (1 ≤ n ≤ 107, 1 ≤ l ≤ r ≤ 107).

Output

Print the integer K — is the answer to the problem.

Examples

Input

1 1 10


Output

10

Input

2 6 9


Output

12

Input

3 1 10


Output

8

Input

3 3 10


Output

2

Note

These are possible progressions for the first test of examples: 

  * 1; 
  * 2; 
  * 3; 
  * 4; 
  * 5; 
  * 6; 
  * 7; 
  * 8; 
  * 9; 
  * 10. 



These are possible progressions for the second test of examples: 

  * 6, 7; 
  * 6, 8; 
  * 6, 9; 
  * 7, 6; 
  * 7, 8; 
  * 7, 9; 
  * 8, 6; 
  * 8, 7; 
  * 8, 9; 
  * 9, 6; 
  * 9, 7; 
  * 9, 8. 



These are possible progressions for the third test of examples: 

  * 1, 2, 4; 
  * 1, 3, 9; 
  * 2, 4, 8; 
  * 4, 2, 1; 
  * 4, 6, 9; 
  * 8, 4, 2; 
  * 9, 3, 1; 
  * 9, 6, 4. 



These are possible progressions for the fourth test of examples: 

  * 4, 6, 9; 
  * 9, 6, 4. 

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        int calc(int n, int l, int r, int mau) {
            int re = 0;
            int mm = IntMath.pow(mau, n - 1);
            int bd = (l - 1) / mm + 1;
            for (; bd * mm <= r; bd++) {
                for (int tu = mau + 1; ; tu++) {
                    long t = 1;
                    boolean ok = true;
                    for (int i = 0; i < n - 1; ++i) {
                        t *= tu;
                        if (t * bd > r) {
                            ok = false;
                            break;
                        }
                    }
                    if (!ok) break;
                    if (IntMath.gcd(mau, tu) == 1) re++;
                }
            }

            return re;
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int l = in.nextInt();
            int r = in.nextInt();
            if (n == 1) {
                out.println(r - l + 1);
            } else if (n == 2) {
                out.println(1l * (r - l) * (r - l + 1));
            } else {
                long re = 0;
                for (int d = 1; ; d++) {
                    boolean ok = true;
                    long t = 1;

                    for (int i = 0; i < n - 1; ++i) {
                        t *= d;
                        if (t > r) {
                            ok = false;
                            break;
                        }
                    }
                    if (!ok) break;
                    re += calc(n, l, r, d);
                }
                out.println(2 * re);
            }
        }

    }

    static final class IntMath {
        public static int pow(int b, int k) {
            MathPreconditions.checkNonNegative("exponent", k);
            switch (b) {
                case 0:
                    return (k == 0) ? 1 : 0;
                case 1:
                    return 1;
                case (-1):
                    return ((k & 1) == 0) ? 1 : -1;
                case 2:
                    return (k < Integer.SIZE) ? (1 << k) : 0;
                case (-2):
                    if (k < Integer.SIZE) {
                        return ((k & 1) == 0) ? (1 << k) : -(1 << k);
                    } else {
                        return 0;
                    }
                default:
                    // continue below to handle the general case
            }
            for (int accum = 1; ; k >>= 1) {
                switch (k) {
                    case 0:
                        return accum;
                    case 1:
                        return b * accum;
                    default:
                        accum *= ((k & 1) == 0) ? 1 : b;
                        b *= b;
                }
            }
        }

        public static int gcd(int a, int b) {
    /*
     * The reason we require both arguments to be >= 0 is because otherwise, what do you return on
     * gcd(0, Integer.MIN_VALUE)? BigInteger.gcd would return positive 2^31, but positive 2^31 isn't
     * an int.
     */
            MathPreconditions.checkNonNegative("a", a);
            MathPreconditions.checkNonNegative("b", b);
            if (a == 0) {
                // 0 % b == 0, so b divides a, but the converse doesn't hold.
                // BigInteger.gcd is consistent with this decision.
                return b;
            } else if (b == 0) {
                return a; // similar logic
            }
    /*
     * Uses the binary GCD algorithm; see http://en.wikipedia.org/wiki/Binary_GCD_algorithm. This is
     * >40% faster than the Euclidean algorithm in benchmarks.
     */
            int aTwos = Integer.numberOfTrailingZeros(a);
            a >>= aTwos; // divide out all 2s
            int bTwos = Integer.numberOfTrailingZeros(b);
            b >>= bTwos; // divide out all 2s
            while (a != b) { // both a, b are odd
                // The key to the binary GCD algorithm is as follows:
                // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
                // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

                // We bend over backwards to avoid branching, adapting a technique from
                // http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax

                int delta = a - b; // can't overflow, since a and b are nonnegative

                int minDeltaOrZero = delta & (delta >> (Integer.SIZE - 1));
                // equivalent to Math.min(delta, 0)

                a = delta - minDeltaOrZero - minDeltaOrZero; // sets a to Math.abs(a - b)
                // a is now nonnegative and even

                b += minDeltaOrZero; // sets b to min(old a, b)
                a >>= Integer.numberOfTrailingZeros(a); // divide out all 2s, since 2 doesn't divide b
            }
            return a << Math.min(aTwos, bTwos);
        }

        private IntMath() {
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public static boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }

    static final class MathPreconditions {
        static int checkNonNegative(String role, int x) {
            if (x < 0) {
                throw new IllegalArgumentException(role + " (" + x + ") must be >= 0");
            }
            return x;
        }

        private MathPreconditions() {
        }

    }
}