N people are waiting in a single line in front of the Takahashi Store. The cash on hand of the i-th person from the front of the line is a positive integer A_i.

Mr. Takahashi, the shop owner, has decided on the following scheme: He picks a product, sets a positive integer P indicating its price, and shows this product to customers in order, starting from the front of the line. This step is repeated as described below.

At each step, when a product is shown to a customer, if price P is equal to or less than the cash held by that customer at the time, the customer buys the product and Mr. Takahashi ends the current step. That is, the cash held by the first customer in line with cash equal to or greater than P decreases by P, and the next step begins.

Mr. Takahashi can set the value of positive integer P independently at each step.

He would like to sell as many products as possible. However, if a customer were to end up with 0 cash on hand after a purchase, that person would not have the fare to go home. Customers not being able to go home would be a problem for Mr. Takahashi, so he does not want anyone to end up with 0 cash.

Help out Mr. Takahashi by writing a program that determines the maximum number of products he can sell, when the initial cash in possession of each customer is given.

Constraints

* 1 ≦ | N | ≦ 100000
* 1 ≦ A_i ≦ 10^9(1 ≦ i ≦ N)
* All inputs are integers.

Input

Inputs are provided from Standard Inputs in the following form.


N
A_1
:
A_N


Output

Output an integer representing the maximum number of products Mr. Takahashi can sell.

Examples

Input

3
3
2
5


Output

3


Input

15
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9


Output

18
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;

public class Main {
	int N;
	int[] A;

	public void solve() {
		N = nextInt();
		A = new int[N];

		for(int i = 0;i < N;i++){
			A[i] = nextInt();
		}

		long max = 1L;
		long ans = (A[0] - 1);//最初の要素を1にし、A[0] - 1をカウント

		for(int i = 1;i < N;i++){
			if(A[i] > max + 1){
				//A[i] - 1は1は残さないといけないため
				//1以上を残してmax+1ずつ減らしていく
				ans += (A[i] - 1) / (max + 1);
			}else if(A[i] == max + 1){
				//A[i]はmax+1で減らすと0になってしまうので、スルー
				//次の要素からはmax = A[i] + 1にしないといけない
				max++;
			}
		}
		out.println(ans);
	}

	public static void main(String[] args) {
		out.flush();
		new Main().solve();
		out.close();
	}

	/* Input */
	private static final InputStream in = System.in;
	private static final PrintWriter out = new PrintWriter(System.out);
	private final byte[] buffer = new byte[2048];
	private int p = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (p < buflen)
			return true;
		p = 0;
		try {
			buflen = in.read(buffer);
		} catch (IOException e) {
			e.printStackTrace();
		}
		if (buflen <= 0)
			return false;
		return true;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrint(buffer[p])) {
			p++;
		}
		return hasNextByte();
	}

	private boolean isPrint(int ch) {
		if (ch >= '!' && ch <= '~')
			return true;
		return false;
	}

	private int nextByte() {
		if (!hasNextByte())
			return -1;
		return buffer[p++];
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = -1;
		while (isPrint((b = nextByte()))) {
			sb.appendCodePoint(b);
		}
		return sb.toString();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}