Given is an integer S. Find how many sequences there are whose terms are all integers greater than or equal to 3, and whose sum is equal to S. The answer can be very large, so output it modulo 10^9 + 7.

Constraints

* 1 \leq S \leq 2000
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


S


Output

Print the answer.

Examples

Input

7


Output

3


Input

2


Output

0


Input

1729


Output

294867501

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
    private static FastReader fr = new FastReader();
    private static PrintWriter out=new PrintWriter(System.out);

    static long binomialCoeff(long n, long k, long MOD)
    {
        long res = 1;

        // Since C(n, k) = C(n, n-k)
        if (k > n - k)
            k = n - k;

        // Calculate value of
        // [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
        for (int i = 0; i < k; ++i) {
            res *= (n - i);
            res /= (i + 1);
            res %= MOD;
        }

        return res;
    }

    public static void main(String[] args) throws IOException {
        StringBuilder sb = new StringBuilder();
        // code goes here
        int a = fr.nextInt();
        int max = a/3;
        long MOD = 1000000007;
        long ways = 0;
        for(int i = 1; i <= max; i++){
            int req = a - i*3;
            ways += nCrModPFermat(req + i - 1, req, MOD);
            ways %= MOD;
        }
        sb.append(ways);
        System.out.print(sb.toString());
    }

    static long power(long x, long y, long p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, long p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r,
                             long p)
    {

        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        long[] fac = new long[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;

        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
                % p;
    }

    static class FastReader{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");

        public String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] nextIntArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public long[] nextLongArray(int n) {
            long[] a=new long[n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
    }

    static class Pair<A, B>{
        A first;
        B second;
        public Pair(A first, B second){
            this.first = first;
            this.second = second;
        }
    }
}
