Problem: Let a be a matrix of size r × c containing positive integers, not necessarily distinct. Rows of the matrix are numbered from 1 to r, columns are numbered from 1 to c. We can construct an array b consisting of r + c integers as follows: for each i ∈ [1, r], let b_i be the greatest common divisor of integers in the i-th row, and for each j ∈ [1, c] let b_{r+j} be the greatest common divisor of integers in the j-th column. 

We call the matrix diverse if all r + c numbers b_k (k ∈ [1, r + c]) are pairwise distinct. 

The magnitude of a matrix equals to the maximum of b_k.

For example, suppose we have the following matrix:

\begin{pmatrix} 2 & 9 & 7\\\ 4 & 144 & 84 \end{pmatrix} 

We construct the array b:

  1. b_1 is the greatest common divisor of 2, 9, and 7, that is 1; 
  2. b_2 is the greatest common divisor of 4, 144, and 84, that is 4; 
  3. b_3 is the greatest common divisor of 2 and 4, that is 2; 
  4. b_4 is the greatest common divisor of 9 and 144, that is 9; 
  5. b_5 is the greatest common divisor of 7 and 84, that is 7. 



So b = [1, 4, 2, 9, 7]. All values in this array are distinct, so the matrix is diverse. The magnitude is equal to 9.

For a given r and c, find a diverse matrix that minimises the magnitude. If there are multiple solutions, you may output any of them. If there are no solutions, output a single integer 0. 

Input

The only line in the input contains two space separated integers r and c (1 ≤ r,c ≤ 500) — the number of rows and the number of columns of the matrix to be found.

Output

If there is no solution, output a single integer 0.

Otherwise, output r rows. The i-th of them should contain c space-separated integers, the j-th of which is a_{i,j} — the positive integer in the i-th row and j-th column of a diverse matrix minimizing the magnitude.

Furthermore, it must hold that 1 ≤ a_{i,j} ≤ 10^9. It can be shown that if a solution exists, there is also a solution with this additional constraint (still having minimum possible magnitude).

Examples

Input


2 2


Output


4 12
2 9

Input


1 1


Output


0

Note

In the first example, the GCDs of rows are b_1 = 4 and b_2 = 1, and the GCDs of columns are b_3 = 2 and b_4 = 3. All GCDs are pairwise distinct and the maximum of them is 4. Since the GCDs have to be distinct and at least 1, it is clear that there are no diverse matrices of size 2 × 2 with magnitude smaller than 4.

In the second example, no matter what a_{1,1} is, b_1 = b_2 will always hold, so there are no diverse matrices.

Solution (in PYTHON):

Solution:
"""
    Satwik_Tiwari ;) .
    28 june , 2020  - Sunday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def bs(a,l,h,x):
    while(l<h):
        # print(l,h)
        mid = (l+h)//2
        if(a[mid] == x):
            return mid
        if(a[mid] < x):
            l = mid+1
        else:
            h = mid
    return l

def sieve(a): #O(n loglogn) nearly linear
    #all odd mark 1
    for i in range(3,((10**6)+1),2):
        a[i] = 1
    #marking multiples of i form i*i 0. they are nt prime
    for i in range(3,((10**6)+1),2):
        for j in range(i*i,((10**6)+1),i):
            a[j] = 0
    a[2] = 1 #special left case
    return (a)


def bfs(g,st):
    visited = [0]*(len(g))
    visited[st] = 1
    queue = []
    queue.append(st)
    new = []
    while(len(queue) != 0):
        s = queue.pop()
        new.append(s)
        for i in g[s]:
            if(visited[i] == 0):
                visited[i] = 1
                queue.append(i)
    return new

def solve():
    n,m = sep()
    if(n==1 and m==1):
        print(0)
        return
    if(m==1):
        val = 2
        for i in range(0,n):
            print(val)
            val+=1
        return
    
    grid = [[0]*(m+1) for i in range(n+1)]

    val = 2
    for i in range(1,m+1):
        grid[0][i] = val
        val+=1
    for i in range(1,n+1):
        if(i==1):
            grid[i][0] = 1
            continue
        grid[i][0] = val
        val+=1

    for i in range(1,n+1):
        for j in range(1,m+1):
            grid[i][j] = grid[i][0]*grid[0][j]
    # print(grid)

    for i in range(1,n+1):
        print(*grid[i][1:])




testcase(1)
# testcase(int(inp()))