Problem: Omkar and Akmar are playing a game on a circular board with n (2 ≤ n ≤ 10^6) cells. The cells are numbered from 1 to n so that for each i (1 ≤ i ≤ n-1) cell i is adjacent to cell i+1 and cell 1 is adjacent to cell n. Initially, each cell is empty.

Omkar and Akmar take turns placing either an A or a B on the board, with Akmar going first. The letter must be placed on an empty cell. In addition, the letter cannot be placed adjacent to a cell containing the same letter. 

A player loses when it is their turn and there are no more valid moves.

Output the number of possible distinct games where both players play optimally modulo 10^9+7. Note that we only consider games where some player has lost and there are no more valid moves.

Two games are considered distinct if the number of turns is different or for some turn, the letter or cell number that the letter is placed on were different.

A move is considered optimal if the move maximizes the player's chance of winning, assuming the other player plays optimally as well. More formally, if the player who has to move has a winning strategy, they have to make a move after which they will still have a winning strategy. If they do not, they can make any move.

Input

The only line will contain an integer n (2 ≤ n ≤ 10^6) — the number of cells on the board.

Output

Output a single integer — the number of possible distinct games where both players play optimally modulo 10^9+7.

Examples

Input


2


Output


4


Input


69420


Output


629909355


Input


42069


Output


675837193

Note

For the first sample case, the first player has 4 possible moves. No matter what the first player plays, the second player only has 1 possible move, so there are 4 possible games.

Solution (in CPP):

Solution:
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ld = long double;
#define X first
#define Y second

const ll mod = 1000000007;

ll gcd(ll a, ll b)
{
    a = abs(a);
    b = abs(b);
    if (a==0 or b==0) return a+b;
    return gcd(b,a%b);
}


ll sum(ll a, ll b)
{
    a += b;
    a %= mod;
    return a;
}

ll sub(ll a, ll b)
{
    a += mod-b;
    a %= mod;
    return a;
}

inline ll mul(ll a, ll b)
{
    if (a*b<mod) return a*b;
    return a*b%mod;
}

inline ll pew(ll a, ll b)
{
    ll res = 1;
    while (b)
    {
        if (b&1) res = mul(res,a);
        a = mul(a,a);
        b /= 2;
    }
    return res;
}



/*const int MOD = 998244353;
const int root = 31;
const int root_1 = 128805723;
const int root_pw = 1<<23;
const int root = 31;
const int MOD = 469762049;
const int root_1 = 15658735;
const int root_pw = 1<<26;*/


/*void fft (vector<int> & a, bool invert) {
    int n = (int) a.size();

    for (int i=1, j=0; i<n; ++i) {
        int bit = n >> 1;
        for (; j>=bit; bit>>=1)
            j -= bit;
        j += bit;
        if (i < j)
            swap (a[i], a[j]);
    }

    for (int len=2; len<=n; len<<=1) {
        int wlen = invert ? root_1 : root;
        for (int i=len; i<root_pw; i<<=1)
            wlen = int (wlen * 1ll * wlen % mod);
        for (int i=0; i<n; i+=len) {
            int w = 1;
            for (int j=0; j<len/2; ++j) {
                int u = a[i+j],  v = int (a[i+j+len/2] * 1ll * w % mod);
                a[i+j] = u+v < mod ? u+v : u+v-mod;
                a[i+j+len/2] = u-v >= 0 ? u-v : u-v+mod;
                w = int (w * 1ll * wlen % mod);
            }
        }
    }
    if (invert) {
        int nrev = pew(n, mod-2);
        for (int i=0; i<n; ++i)
            a[i] = int (a[i] * 1ll * nrev % mod);
    }
}*/


int parent[600001];
int rnk[600001];

void make_set (int v) {
    parent[v] = v;
    rnk[v] = 0;
}

int find_set (int v) {
    if (v == parent[v])
        return v;
    return parent[v] = find_set(parent[v]);
}

void union_sets (int a, int b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
        if (rnk[a] < rnk[b])
            swap(a, b);
        parent[b] = a;
        if (rnk[a] == rnk[b])
            ++rnk[a];
    }
}

//ll t[1200000];
//ll del[1200000];

/*void build()
{
    for (int i=0;i<1200000;i++)
    {
        t[i] = 0;
        //del[i] = 0;
    }
}*/

/*void push(int v, int tl, int tr, int tm)
{
    t[2*v].X += del[v];
    t[2*v+1].X += del[v];
    del[2*v] += del[v];
    del[2*v+1] += del[v];
    del[v] = 0;
}*/

/*void update (int v, int tl, int tr, int l, int r, ll val) {
    if (tl==l and tr==r)
    {
        //del[v] += val;

        //t[v] += val*(tr-tl+1);
        return;
    }
    int tm = (tl + tr) / 2;
    //if (del[v]!=0) push(v,tl,tr,tm);
    if (r<=tm)
    {
        update(2*v,tl,tm,l,r,val);
        t[v] = t[2*v]+t[2*v+1];
        return;
    }
    if (l>tm)
    {
        update(2*v+1,tm+1,tr,l,r,val);
        t[v] = t[2*v]+t[2*v+1];
        return;
    }
    update(2*v,tl,tm,l,tm,val);
    update(2*v+1,tm+1,tr,tm+1,r,val);
    t[v] = min(t[2*v],t[2*v+1]);
}*/

/*
void update (int v, int tl, int tr, int l, int r, ll val) {
    if (tl==l and tr==r)
    {
        //del[v] += val;
        t[v] += val;
        return;
    }
    int tm = (tl+tr)/2;
    //push(v,tl,tr,tm);
    if (r<=tm)
    {
        update(2*v,tl,tm,l,r,val);
        return;
    }
    if (l>tm)
    {
        update(2*v+1,tm+1,tr,l,r,val);
        return;
    }
    update(2*v,tl,tm,l,tm,val);
    update(2*v+1,tm+1,tr,tm+1,r,val);
    //t[v] = min(t[2*v],t[2*v+1]);
}


ll get (int v, int tl, int tr, int p) {
    //if (del[v]!=0 and tl!=tr) push(v,tl,tr,tm);
    if (tl==tr)
    {
        return t[v];
    }
    int tm = (tl+tr)/2;
    //push(v,tl,tr,tm);
    if (p<=tm) return t[v]+get(2*v,tl,tm,p);
    else return t[v]+get(2*v+1,tm+1,tr,p);
}*/





int main()
{
    ios_base::sync_with_stdio(false);
    ll n;
    cin >> n;
    if (n<=2)
    {
        cout << n*2 << endl;
        return 0;
    }
    vector<ll> f(n+1), obrf(n+1);
    f[0] = 1;
    obrf[0] = 1;
    for (ll i=1;i<=n;i++)
    {
        f[i] = f[i-1]*i%mod;
        obrf[i] = pew(f[i],mod-2);
    }
    ll ans = 0;
    for (ll k=0;k*2<=n;k++)
    {
        if (n>4*k) continue;
        ans = (ans+f[2*k]*obrf[n-2*k]%mod*obrf[4*k-n]%mod*f[2*k-1]%mod*2*n)%mod;
    }
    cout << ans << endl;
}