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:
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define x first
#define y second
#define LL long long
#define LD long double
#define Complex complex<double>
#define y1 ___y1
#define next ___next
#define left ___left
#define right ___right
#define pii pair<int, int>
#define pll pair<LL, LL>
#define pdd pair<double, double>
#define vi vector<int>
#define vl vector<LL>
#define vii vector<pii>
#define vll vector<pll>
#define PQ priority_queue
#define ffo fflush(stdout)
#define gch getchar()
#define e (int)(1e9 + 7)
#define MOD (int)(1e9 + 7)
#define e2 (double)2.71828182846
#define pi (double)3.14159265358979
#define ri(X) scanf("%d", &(X))
#define rii(X, Y) scanf("%d%d", &(X), &(Y))
#define riii(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define rl(X) scanf("%lld", &(X))
#define rll(X, Y) scanf("%lld%lld", &(X), &(Y))
#define rlll(X, Y, Z) scanf("%lld%lld%lld", &(X), &(Y), &(Z))
#define printi(X) printf("%d", (int)(X))
#define printii(X, Y) printf("%d %d", (int)(X), (int)(Y))
#define printiii(X, Y, Z) printf("%d %d %d", (int)(X), (int)(Y), (int)(Z))
#define printl(X) printf("%lld", (LL)(X))
#define printll(X, Y) printf("%lld %lld", (LL)(X), (LL)(Y))
#define printlll(X, Y, Z) printf("%lld %lld %lld", (LL)(X), (LL)(Y), (LL)(Z))
#define enter printf("\n")
#define space printf(" ")
struct node {
    int x, y;
    bool operator < (const node &T) const {
        return x < T.x;
    }
};
struct matrix {
    int sz;
    LL a[105][105];
    matrix& operator = (const matrix &T) {
        sz = T.sz;
        for (int i = 1; i <= sz; i++) {
            for (int j = 1; j <= sz; j++) {
                a[i][j] = T.a[i][j];
            }
        }
        return *this;
    }
    matrix operator + (const matrix &T) const {
        matrix ans;
        ans.sz = sz;
        for (int i = 1; i <= sz; i++) {
            for (int j = 1; j <= sz; j++) {
                ans.a[i][j] = a[i][j] + T.a[i][j];
                ans.a[i][j] %= e;
            }
        }
        return ans;
    }
    matrix operator * (const matrix &T) const {
        matrix ans;
        ans.sz = sz;
        for (int i = 1; i <= sz; i++) {
            for (int j = 1; j <= sz; j++) {
                ans.a[i][j] = 0;
                for (int k = 1; k <= sz; k++) {
                    ans.a[i][j] += a[i][k] * T.a[k][j];
                    ans.a[i][j] %= e;
                }
            }
        }
        return ans;
    }
    matrix I() {
        matrix ans;
        ans.sz = sz;
        for (int i = 1; i <= sz; i++) {
            for (int j = 1; j <= sz; j++) {
                if (i == j) ans.a[i][j] = 1;
                else ans.a[i][j] = 0;
            }
        }
        return ans;
    }
    matrix power(LL p) {
        if (p == 0) return I();
        if (p == 1) return *this;
        matrix tmp = power(p / 2);
        if (p % 2) tmp = tmp * tmp * (*this);
        else tmp = tmp * tmp;
        return tmp;
    }
    void print() {
        for (int i = 1; i <= sz; i++) {
            for (int j = 1; j <= sz; j++) {
                printf(j == sz? "%lld\n" : "%lld ", a[i][j]);
            }
        }
        return;
    }
};
struct polynomial {
    vector<Complex> v;
    int sz;
    polynomial operator * (const polynomial &T) const {
        polynomial ans;
        if (sz >= T.sz) {
            ans = *this;
            for (int i = 0; i < T.sz; i++) {
                ans.v[i] *= T.v[i];
            }
        }
        else {
            ans = T;
            for (int i = 0; i < sz; i++) {
                ans.v[i] *= v[i];
            }
        }
        return ans;
    }
};
polynomial FFT(polynomial a) {
    if (a.sz <= 1) return a;
    polynomial even, odd;
    even.sz = odd.sz = 0;
    for (int i = 0; i < a.sz; i += 2) {
        even.v.pb(a.v[i]);
        even.sz++;
    }
    for (int i = 1; i < a.sz; i += 2) {
        odd.v.pb(a.v[i]);
        odd.sz++;
    }
    polynomial Even = FFT(even);
    polynomial Odd = FFT(odd);
    polynomial ans;
    ans.sz = a.sz;
    for (int i = 0; i < ans.sz; i++) {
        ans.v.pb(0);
    }
    for (int i = 0; i < a.sz / 2; i++) {
        Complex w = polar(1.0, -2 * pi * i / a.sz) * Odd.v[i];
        ans.v[i] = Even.v[i] + w;
        ans.v[i + a.sz / 2] = Even.v[i] - w;
    }
    return ans;
}
polynomial IFFT(polynomial a) {
    for (int i = 0; i < a.sz; i++) {
        a.v[i] = conj(a.v[i]);
    }
    polynomial ans = FFT(a);
    for (int i = 0; i < a.sz; i++) {
        ans.v[i] = conj(ans.v[i]);
        ans.v[i] /= ans.sz;
    }
    return ans;
}
struct polygon {
    int n;
    pii a[1005];
    double area() {
        double d1 = 0, d2 = 0;
        for (int i = 0; i < n; i++) {
            d1 += a[i].x * a[(i + 1) % n].y;
            d2 += a[(i + 1) % n].x * a[i].y;
        }
        return (d1 - d2) / 2;
    }
};
LL power(LL a, LL p) {
    if (p == 0) return 1;
    if (p == 1) return a;
    LL tmp = power(a, p / 2);
    if (p % 2) tmp = tmp * tmp % MOD * a;
    else tmp *= tmp;
    return tmp % MOD;
}
vl inverse(int n) {
    vl ans(n + 1);
    ans[0] = ans[1] = 1;
    for (int i = 2; i <= n; i++) {
        ans[i] = ans[MOD % i] * (MOD - MOD / i) % MOD;
    }
    return ans;
}
vl factorial(int n) {
    vl ans(n + 1);
    ans[0] = ans[1] = 1;
    for (int i = 2; i <= n; i++) {
        ans[i] = ans[i - 1] * i % MOD;
    }
    return ans;
}
vl factorialInverse(int n) {
    vl inv = inverse(n);
    vl ans(n + 1);
    ans[0] = ans[1] = 1;
    for (int i = 2; i <= n; i++) {
        ans[i] = inv[i] * ans[i - 1] % MOD;
    }
    return ans;
}
int T;
int n;
vl fac, inv;
LL ans;

LL ncr(int n, int r) {
    return fac[n] * inv[r] % MOD * inv[n - r] % MOD;
}

void solve(int CASE) {
    ri(n);
    fac = factorial(n);
    inv = factorialInverse(n);
    for (int i = (n + 1) / 2; i <= n; i++) {
        if (i % 2 != 0) continue;
        LL add = ncr(i, n - i) + (i == n? 0 : ncr(i - 1, n - i - 1));
        add = add % MOD * fac[i] % MOD;
        ans = ans + add % MOD;
    }
    ans = ans * 2 % MOD;
    printl(ans);
    enter;
    return;
}

int main() {
    // ios::sync_with_stdio(false);
    // cin.tie(NULL), cout.tie(NULL);
    T = 1;
    for (int _ = 1; _ <= T; _++) {
        solve(_);
    }
    return 0;
}