Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1 × n in size, when viewed from above. This rectangle is divided into n equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.

Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1 × 5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:

<image>

As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him. 

Input

The first line contains a positive integer n (1 ≤ n ≤ 1000). The second line contains n positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.

Output

Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.

Examples

Input

1
2


Output

1


Input

5
1 2 1 2 1


Output

3


Input

8
1 2 1 1 1 3 3 4


Output

6
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by adi0311(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import heapq
from math import *
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(var)
def l(): return list(map(int, data().split()))
def sl(): return list(map(str, data().split()))
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


def left(arr, i):
    res = 1
    j = i+1
    while j < n and arr[j] <= arr[j-1]:
        res += 1
        j += 1
    return res


def right(arr, i):
    res = 1
    j = i-1
    while j >= 0 and arr[j] <= arr[j+1]:
        res += 1
        j -= 1
    return res


n = int(data())
arr = l()
answer = 0
for i in range(n):
    if i == 0:
        res = left(arr, i)
    elif i == n-1:
        res = right(arr, i)
    else:
        res = left(arr, i)+right(arr, i)-1
    answer = max(answer, res)
print(answer)
