def find_peak(arr, n):
	def find_peak_util(arr, low, high, n): 
		mid = low + (high - low)/2
		mid = int(mid) 
		if ((mid == 0 or arr[mid - 1] <= arr[mid]) and
			(mid == n - 1 or arr[mid + 1] <= arr[mid])): 
			return mid 
		elif (mid > 0 and arr[mid - 1] > arr[mid]): 
			return find_peak_util(arr, low, (mid - 1), n) 
		else: 
			return find_peak_util(arr, (mid + 1), high, n) 
	 
	return find_peak_util(arr, 0, n - 1, n) 
def min_cost(cost, m, n):
	R = 3
	C = 3
	 
	tc = [[0 for x in range(C)] for x in range(R)] 
	tc[0][0] = cost[0][0] 
	for i in range(1, m+1): 
		tc[i][0] = tc[i-1][0] + cost[i][0] 
	for j in range(1, n+1): 
		tc[0][j] = tc[0][j-1] + cost[0][j] 
	for i in range(1, m+1): 
		for j in range(1, n+1): 
			tc[i][j] = min(tc[i-1][j-1], tc[i-1][j], tc[i][j-1]) + cost[i][j] 
	return tc[m][n]