There are N slimes standing on a number line. The i-th slime from the left is at position x_i.

It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.

Niwango will perform N-1 operations. The i-th operation consists of the following procedures:

* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.



Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.

Input

Input is given from Standard Input in the following format:


N
x_1 x_2 \ldots x_N


Output

Print the answer.

Examples

Input

3
1 2 3


Output

5


Input

12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932


Output

750927044
import java.util.*;
import java.io.*;

public class Main {
    static long mod = (long) 1e9 + 7;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt();
        long [] x = new long[n + 1];
        for (int i = 1; i <= n; i++) x[i] = sc.nextInt();
        long [] pref = new long[n];
        pref[1] = 1;
        for (int i = 2; i < n; i++) pref[i] = (pref[i - 1] + inv(i)) % mod;
        long ans = 0;
        for (int i = 1; i < n; i++) {
            ans = (ans + pref[i] * (x[i + 1] - x[i]) % mod) % mod;
        }
        fac = new long[n + 3];
        initFac(n + 1);
        ans = (ans * fac[n - 1]) % mod;
        out.println(ans);
        out.close();
    }

    static long[] fac;

    static void initFac(long n) {
        fac[0] = 1;
        for (int i = 1; i <= n; i++) {
            fac[i] = (fac[i - 1]  * i) % mod;
        }
    }

    static long exponentiation(long base, long exp)
    {
        long t = 1L;
        while (exp > 0) {

            // for cases where exponent
            // is not an even value
            if (exp % 2 != 0)
                t = (t * base) % mod;

            base = (base * base) % mod;
            exp /= 2;
        }
        return t % mod;
    }

    static long nck(int n, int k) {
        if (n < k)
            return 0;
        long den = inv((int) (fac[k] * fac[n - k] % mod));
        return fac[n] * den % mod;
    }

    static long pow(long b, long e) {
        long ans = 1;
        while (e > 0) {
            if (e % 2 == 1)
                ans = ans * b % mod;
            e >>= 1;
            b = b * b % mod;
        }
        return ans;
    }

    static long inv(long x) {
        return pow(x, mod - 2);
    }



    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}