Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) he makes ai money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence ai. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Examples

Input

6
2 2 1 3 4 1


Output

3

Input

3
2 2 9


Output

3

Note

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
import math
from decimal import Decimal
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
 
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
 
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def dva():
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
 
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m
 

def ist(x):
	d = 1
	while d < x:
		d *= 2
	return d == x


n, a = na()
k = 1
s = set()
prev = a[0]
for i in range(1, n):
	if a[i] >= prev:
		k += 1
		prev = a[i]
	else:
		prev = a[i]
		s.add(k)
		k = 1
if len(a) == 1:
	print(1)
	exit()
if a[i] >= a[i - 1]:
	s.add(k)
print(max(s))
