Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.

You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.

Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Examples

Input

4
0 1 0 10


Output

1


Input

2
0 0


Output

2


Input

10
1 1 3 0 0 0 2 1 0 3


Output

3

Note

In first sample the last person kills everyone in front of him.
import java.util.*;

public class Codeforces_446_2 {
	
	public static void main(String[] args)
	{
		
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		
		Stack<Integer> s = new Stack<Integer>();
		
		for(int i=0;i<n;i++)
		{
			int x = input.nextInt();
			s.push(x);
		}
		
		int ans = 0;
		while(!s.isEmpty())
		{
			int x = s.pop();
			ans++;
			
			while(x>0 && !s.isEmpty())
			{
				int y = s.pop();
				if(y>(x-1))
					x = y;
				else
					x--;
			}
		}
		System.out.println(ans);
		
//		int[] a = new int[n];
//		int[] dp = new int[n+1];
//		for(int i=0;i<n;i++)
//			a[i] = input.nextInt();
//		
//		dp[0] = 1;
//		for(int i=1;i<=n;i++)
//		{
//			int x = a[i-1];
//			if((i-x)>=0)
//			{
//				if(x==0)
//					dp[i] = 1 + dp[i-1];
//				else
//					dp[i] = dp[i-x];
//			}
//			else
//				dp[i] = 1;
//			
//			if(dp[i]>i)
//				dp[i] = i;
//		}
		
//		for(int i=1;i<=n;i++)
//			System.out.print(dp[i] + " ");
//		
//		System.out.println();
		
//		System.out.println(dp[n]);
	}

}
