#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <iterator>
#include <cassert>
#include <numeric>
#include <functional>
#include <ctime>
#pragma warning(disable:4996) 

//#define ATCODER
#ifdef ATCODER
#include <atcoder/all>
#endif

typedef long long ll;
typedef unsigned long long ull;
#define LINF  9223300000000000000
#define LINF2 1223300000000000000
#define LINF3 1000000000000
#define INF 2140000000
const long long MOD = 1000000007;
//const long long MOD = 998244353;

using namespace std;
#ifdef ATCODER
using namespace atcoder;
#endif

ll mpow(ll x, ll n) { //x^n(mod M)
    ll ans = 1;
    while (n != 0) {
        if (n & 1) ans = ans * x % MOD;
        x = x * x % MOD;
        n = n >> 1;
    }
    return ans;
}

ll minv(ll x) {
    return mpow(x, MOD - 2);
}

void solve()
{
    ll n;
    scanf("%lld", &n);
    if (n == 0) {
        printf("1\n");
    }
    else if (n == 1) {
        printf("12\n");
    }
    else if (n == 2) {
        printf("65\n");
    }
    else if(n%2) {
        ll ans0 = ((3 * n + 1) % MOD)*((3 * n) % MOD)%MOD*minv(2) % MOD;
        ll ans1 = ((n + 1) / 2) * ((n - 1) / 2) % MOD;
        ll ans = (ans0 - ans1 + MOD) * 4 % MOD;
        printf("%lld\n", ans);
    }
    else {
        ll ans0 = ((3 * n / 2 + 1) % MOD)*((3 * n / 2 + 1) % MOD) % MOD;
        ll ans1 = ((3 * n / 2) % MOD)*((3 * n / 2) % MOD) % MOD;
        ll ans2 = (n / 2) * (n / 2) % MOD;
        ll ans3 = (n * 3 / 2 + 1) % MOD;
        ll ans = ((ans0 + ans1- ans2 - ans3+ MOD*2) * 4 + 1) % MOD;
        printf("%lld\n", ans);
    }
    return;
}


int main()
{
#if 1
    solve();
#else
    int T, t;
    scanf("%d", &T);
    for (t = 0; t < T; t++) {
        printf("Case #%d: ", t + 1);
        solve();
    }
#endif
    return 0;
}