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
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
const int nMax=1e5, MOD=1e9+7;
ULL inv[nMax], Fact[nMax], invF[nMax];
ULL pw(ULL a, ULL n){ ULL b=1; while(n){ if(n&1){ b=(b*a)%MOD; } a=(a*a)%MOD; n>>=1; } return b; }
int main(){
inv[1]=Fact[0]=Fact[1]=invF[0]=invF[1]=1;
for(int i=2; i<nMax; i++){ 
  inv[i]=((MOD-MOD/i)*inv[MOD%i])%MOD;
  Fact[i]=(Fact[i-1]*i)%MOD; invF[i]=(invF[i-1]*inv[i])%MOD;
}
ULL N, ans=0; cin >> N; ULL x[N], p[N-1];
for(int i=0; i<N; i++){ cin >> x[i]; }
p[0]=Fact[N-1];
for(int i=1; i<N-1; i++){ p[i]=(p[i-1]+(Fact[N-1]*pw(i+1, 1e9+5))%MOD)%MOD; }
for(int i=0; i<N-1; i++){ ans=(ans+((x[i+1]-x[i])*p[i])%MOD)%MOD; }
cout << ans << "\n";
}