Problem: Lunar New Year is approaching, and Bob received a gift from his friend recently — a recursive sequence! He loves this sequence very much and wants to play with it.

Let f_1, f_2, …, f_i, … be an infinite sequence of positive integers. Bob knows that for i > k, f_i can be obtained by the following recursive equation:

$$$f_i = \left(f_{i - 1} ^ {b_1} ⋅ f_{i - 2} ^ {b_2} ⋅ ⋅⋅⋅ ⋅ f_{i - k} ^ {b_k}\right) mod p,$$$

which in short is

$$$f_i = \left(∏_{j = 1}^{k} f_{i - j}^{b_j}\right) mod p,$$$

where p = 998 244 353 (a widely-used prime), b_1, b_2, …, b_k are known integer constants, and x mod y denotes the remainder of x divided by y.

Bob lost the values of f_1, f_2, …, f_k, which is extremely troublesome – these are the basis of the sequence! Luckily, Bob remembers the first k - 1 elements of the sequence: f_1 = f_2 = … = f_{k - 1} = 1 and the n-th element: f_n = m. Please find any possible value of f_k. If no solution exists, just tell Bob that it is impossible to recover his favorite sequence, regardless of Bob's sadness.

Input

The first line contains a positive integer k (1 ≤ k ≤ 100), denoting the length of the sequence b_1, b_2, …, b_k.

The second line contains k positive integers b_1, b_2, …, b_k (1 ≤ b_i < p).

The third line contains two positive integers n and m (k < n ≤ 10^9, 1 ≤ m < p), which implies f_n = m.

Output

Output a possible value of f_k, where f_k is a positive integer satisfying 1 ≤ f_k < p. If there are multiple answers, print any of them. If no such f_k makes f_n = m, output -1 instead.

It is easy to show that if there are some possible values of f_k, there must be at least one satisfying 1 ≤ f_k < p.

Examples

Input


3
2 3 5
4 16


Output


4


Input


5
4 7 1 5 6
7 14187219


Output


6


Input


8
2 3 5 6 1 7 9 10
23333 1


Output


1


Input


1
2
88888 66666


Output


-1


Input


3
998244352 998244352 998244352
4 2


Output


-1


Input


10
283 463 213 777 346 201 463 283 102 999
2333333 6263423


Output


382480067

Note

In the first sample, we have f_4 = f_3^2 ⋅ f_2^3 ⋅ f_1^5. Therefore, applying f_3 = 4, we have f_4 = 16. Note that there can be multiple answers.

In the third sample, applying f_7 = 1 makes f_{23333} = 1.

In the fourth sample, no such f_1 makes f_{88888} = 66666. Therefore, we output -1 instead.

Solution (in JAVA):

Solution:
// upsolve with rainboy
import java.io.*;
import java.util.*;

public class CF1106F {
	static final int MD = 998244353;
	static void mult(int[][] aa, int[][] bb, int[][] cc, int k) {
		for (int i = 0; i < k; i++) {
			int[] ci = cc[i];
			for (int j = 0; j < k; j++)
				ci[j] = 0;
			for (int h = 0; h < k; h++) {
				int[] bh = bb[h];
				long aih = aa[i][h];
				for (int j = 0; j < k; j++)
					ci[j] = (int) ((ci[j] + aih * bh[j]) % (MD - 1));
			}
		}
	}
	static void power_(int[][] aa, int[][] pp, int[][] tt, int k, int x) {
		if (x == 0) {
			for (int i = 0; i < k; i++)
				for (int j = 0; j < k; j++)
					pp[i][j] = i == j ? 1 : 0;
			return;
		}
		if (x % 2 == 0) {
			power_(aa, tt, pp, k, x / 2);
			mult(tt, tt, pp, k);
		} else {
			power_(aa, pp, tt, k, x / 2);
			mult(pp, pp, tt, k);
			mult(tt, aa, pp, k);
		}
	}
	static int power(int a, int x, int md) {
		if (x == 0)
			return 1;
		int p = power(a, x / 2, md);
		p = (int) ((long) p * p % md);
		if (x % 2 == 1)
			p = (int) ((long) p * a % md);
		return p;
	}
	// https://math.stackexchange.com/questions/124408/finding-a-primitive-root-of-a-prime-number
	static int totient(int n) {
		int t = n;
		for (int a = 2; a * a <= n; a++)
			if (n % a == 0) {
				t = t / a * (a - 1);
				do
					n /= a;
				while (n % a == 0);
			}
		if (n > 1) {
			int a = n;
			t = t / a * (a - 1);
		}
		return t;
	}
	static boolean test(int r, int t, int n) {
		int t_ = t;
		for (int a = 2; a * a <= t; a++)
			if (t % a == 0) {
				if (power(r, t_ / a, n) == 1)
					return false;
				do
					t /= a;
				while (t % a == 0);
			}
		if (t > 1) {
			int a = t;
			if (power(r, t_ / a, n) == 1)
				return false;
		}
		return true;
	}
	static int primroot(int n) {
		int t = totient(n);
		Random rand = new Random();
		while (true) {
			int r = rand.nextInt(n - 2) + 2;	// [2, n - 1]
			if (test(r, t, n))
				return r;
		}
	}
	// https://en.wikipedia.org/wiki/Baby-step_giant-step
	static int dislog(int a, int b) {
		int m = (int) Math.sqrt(MD) + 1;
		HashMap<Integer, Integer> map = new HashMap<>();
		int p = 1;
		for (int j = 0; j < m; j++) {
			map.put(p, j);
			p = (int) ((long) p * a % MD);
		}
		int am = power(power(a, MD - 2, MD), m, MD);
		p = b;
		for (int i = 0, j; i < m; i++) {
			if ((j = map.getOrDefault(p, -1)) != -1)
				return (i * m + j) % (MD - 1);
			p = (int) ((long) p * am % MD);
		}
		return -1;
	}
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	static int x_, y_;
	static int gcd_(int a, int b) {
		if (b == 0) {
			x_ = 1; y_ = 0;
			return a;
		} else {
			int d = gcd_(b, a % b);
			int t = x_ - a / b * y_; x_ = y_; y_ = t;
			return d;
		}
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int k = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] bb = new int[k];
		for (int i = 0; i < k; i++)
			bb[i] = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int[][] aa = new int[k][k];
		for (int i = 0; i < k - 1; i++)
			aa[i][i + 1] = 1;
		for (int j = 0; j < k; j++)
			aa[k - 1][j] = bb[k - 1 - j];
		int[][] pp = new int[k][k];
		int[][] tt = new int[k][k];
		power_(aa, pp, tt, k, n - k);
		int d = gcd_(pp[k - 1][k - 1], MD - 1);
		int r = primroot(MD);
		int z = dislog(r, m);
		int ans = -1;
		if (z % d == 0) {
			if (x_ < 0)
				x_ += MD - 1;
			ans = power(r, (int) ((long) x_ * (z / d) % (MD - 1)), MD);
		}
		System.out.println(ans);
	}
}