Problem: There are n types of coins in Byteland. Conveniently, the denomination of the coin type k divides the denomination of the coin type k + 1, the denomination of the coin type 1 equals 1 tugrick. The ratio of the denominations of coin types k + 1 and k equals ak. It is known that for each x there are at most 20 coin types of denomination x.

Byteasar has bk coins of type k with him, and he needs to pay exactly m tugricks. It is known that Byteasar never has more than 3·105 coins with him. Byteasar want to know how many ways there are to pay exactly m tugricks. Two ways are different if there is an integer k such that the amount of coins of type k differs in these two ways. As all Byteland citizens, Byteasar wants to know the number of ways modulo 109 + 7.

Input

The first line contains single integer n (1 ≤ n ≤ 3·105) — the number of coin types.

The second line contains n - 1 integers a1, a2, ..., an - 1 (1 ≤ ak ≤ 109) — the ratios between the coin types denominations. It is guaranteed that for each x there are at most 20 coin types of denomination x.

The third line contains n non-negative integers b1, b2, ..., bn — the number of coins of each type Byteasar has. It is guaranteed that the sum of these integers doesn't exceed 3·105.

The fourth line contains single integer m (0 ≤ m < 1010000) — the amount in tugricks Byteasar needs to pay.

Output

Print single integer — the number of ways to pay exactly m tugricks modulo 109 + 7.

Examples

Input

1

4
2


Output

1


Input

2
1
4 4
2


Output

3


Input

3
3 3
10 10 10
17


Output

6

Note

In the first example Byteasar has 4 coins of denomination 1, and he has to pay 2 tugricks. There is only one way.

In the second example Byteasar has 4 coins of each of two different types of denomination 1, he has to pay 2 tugricks. There are 3 ways: pay one coin of the first type and one coin of the other, pay two coins of the first type, and pay two coins of the second type.

In the third example the denominations are equal to 1, 3, 9.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class E {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	static final int P = 1_000_000_007;
	
	int[] update(int[] dp, int cnt) {
		int[] ret = new int[dp.length + cnt];
		int sum = 0;
		for (int i = 0; i < ret.length; i++) {
			if (i < dp.length) {
				sum += dp[i];
				if (sum >= P) {
					sum -= P;
				}
			}
			ret[i] = sum;
			if (i >= cnt) {
				sum -= dp[i - cnt];
				if (sum < 0) {
					sum += P;
				}
			}
		}
		return ret;
	}
	
	void solve() throws IOException {
		int n = nextInt();
		int[] a = new int[n - 1];
		for (int i = 0; i < n - 1; i++) {
			a[i] = nextInt();
		}

		int[] b = new int[n];
		for (int i = 0; i < n; i++) {
			b[i] = nextInt();
		}

		int[] dp = { 1 };
		
		BigInteger big = new BigInteger(nextToken());
		
		for (int i = 0; i < n; i++) {
			dp = update(dp, b[i]);
			
			if (i == n - 1) {
				break;
			}
			
			if (a[i] == 1) {
				continue;
			}
			
			BigInteger[] tmp = big.divideAndRemainder(BigInteger.valueOf(a[i]));
			
			int rem = tmp[1].intValue();
			
			if (rem >= dp.length) {
				out.println(0);
				return;
			}
			
			int newLen = (dp.length - 1 - rem) / a[i] + 1;
			int[] newDp = new int[newLen];
			
			for (int ii = rem, j = 0; j < newLen; ii += a[i], j++) {
				newDp[j] = dp[ii];
			}
			
			big = tmp[0];
			dp = newDp;
		}
		
		if (big.compareTo(BigInteger.valueOf(dp.length)) < 0) {
			out.println(dp[big.intValue()]);
		} else {
			out.println(0);
		}
	}

	E() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new E();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}