真香，真好用，只許給出前幾項你就會體驗前所未有的快樂。

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <string>

#include <map>

#include <set>

#include <cassert>

#include <iostream>

using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define SZ(x) ((int)(x).size())

typedef vector<int> VI;

typedef long long ll;

typedef pair<int, int> PII;

const ll mod = 1000000007;

ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a%mod; a = a * a%mod; }return res; }

// head

 

int _;

ll n;

namespace linear_seq {

	const int N = 10010;

	ll res[N], base[N], _c[N], _md[N];

 

	vector<int> Md;

	void mul(ll *a, ll *b, int k) {

		rep(i, 0, k + k) _c[i] = 0;

		rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;

		for (int i = k + k - 1; i >= k; i--) if (_c[i])

			rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;

		rep(i, 0, k) a[i] = _c[i];

	}

	int solve(ll n, VI a, VI b) { // a 係數 b 初值 b[n+1]=a[0]*b[n]+...

								  //        printf("%d\n",SZ(b));

		ll ans = 0, pnt = 0;

		int k = SZ(a);

		assert(SZ(a) == SZ(b));

		rep(i, 0, k) _md[k - 1 - i] = -a[i]; _md[k] = 1;

		Md.clear();

		rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);

		rep(i, 0, k) res[i] = base[i] = 0;

		res[0] = 1;

		while ((1ll << pnt) <= n) pnt++;

		for (int p = pnt; p >= 0; p--) {

			mul(res, res, k);

			if ((n >> p) & 1) {

				for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;

				rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;

			}

		}

		rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;

		if (ans<0) ans += mod;

		return ans;

	}

	VI BM(VI s) {

		VI C(1, 1), B(1, 1);

		int L = 0, m = 1, b = 1;

		rep(n, 0, SZ(s)) {

			ll d = 0;

			rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;

			if (d == 0) ++m;

			else if (2 * L <= n) {

				VI T = C;

				ll c = mod - d * powmod(b, mod - 2) % mod;

				while (SZ(C)<SZ(B) + m) C.pb(0);

				rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;

				L = n + 1 - L; B = T; b = d; m = 1;

			}

			else {

				ll c = mod - d * powmod(b, mod - 2) % mod;

				while (SZ(C)<SZ(B) + m) C.pb(0);

				rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;

				++m;

			}

		}

		return C;

	}

	int gao(VI a, ll n) {

		VI c = BM(a);

		c.erase(c.begin());

		rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;

		return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));

	}

};

 

int main() {

    int t;

    cin >> t;

	while(t--){

		scanf("%lld", &n);

		printf("%d\n", linear_seq::gao(VI{ 3, 9, 20, 46, 106, 244, 560, 1286, 2956, 6794 }, n - 1));//在gao後面儘可能的給出前n項，給的越多，結果越對。

	}

}