Hacker Zhorik wants to decipher two secret messages he intercepted yesterday. Yeah message is a sequence of encrypted blocks, each of them consists of several bytes of information.

Zhorik knows that each of the messages is an archive containing one or more files. Zhorik knows how each of these archives was transferred through the network: if an archive consists of k files of sizes l1, l2, ..., lk bytes, then the i-th file is split to one or more blocks bi, 1, bi, 2, ..., bi, mi (here the total length of the blocks bi, 1 + bi, 2 + ... + bi, mi is equal to the length of the file li), and after that all blocks are transferred through the network, maintaining the order of files in the archive.

Zhorik thinks that the two messages contain the same archive, because their total lengths are equal. However, each file can be split in blocks in different ways in the two messages.

You are given the lengths of blocks in each of the two messages. Help Zhorik to determine what is the maximum number of files could be in the archive, if the Zhorik's assumption is correct.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 105) — the number of blocks in the first and in the second messages.

The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 106) — the length of the blocks that form the first message.

The third line contains m integers y1, y2, ..., ym (1 ≤ yi ≤ 106) — the length of the blocks that form the second message.

It is guaranteed that x1 + ... + xn = y1 + ... + ym. Also, it is guaranteed that x1 + ... + xn ≤ 106.

Output

Print the maximum number of files the intercepted array could consist of.

Examples

Input

7 6
2 5 3 1 11 4 4
7 8 2 4 1 8


Output

3


Input

3 3
1 10 100
1 100 10


Output

2


Input

1 4
4
1 1 1 1


Output

1

Note

In the first example the maximum number of files in the archive is 3. For example, it is possible that in the archive are three files of sizes 2 + 5 = 7, 15 = 3 + 1 + 11 = 8 + 2 + 4 + 1 and 4 + 4 = 8.

In the second example it is possible that the archive contains two files of sizes 1 and 110 = 10 + 100 = 100 + 10. Note that the order of files is kept while transferring archives through the network, so we can't say that there are three files of sizes 1, 10 and 100.

In the third example the only possibility is that the archive contains a single file of size 4.
"""
Author    : co_devil Chirag Garg
Institute : JIIT
"""

from __future__ import division, print_function
import itertools, os, sys, threading
from collections import deque, Counter, OrderedDict, defaultdict
import heapq
from math import ceil,floor,log,sqrt,factorial,pow,pi
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
"""from io import BytesIO, IOBase
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    from builtins import str as __str__
    str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._buffer = BytesIO()
        self._fd = file.fileno()
        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):
        return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
            self.newlines += b.count(b'\n') + (not b)
        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)
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')

def print(*args, **kwargs):
    sep, file = kwargs.pop('sep', b' '), 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', b'\n'))
    if kwargs.pop('flush', False):
        file.flush()

"""


def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int,input().split())
def li(): return list(mi())


abc = 'abcdefghijklmnopqrstuvwxyz'
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}
mod = 1000000007
dx, dy = [-1, 1, 0, 0], [0, 0, 1, -1]
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
# For getting input from input.txt file
# sys.stdin = open('input.txt', 'r')

# Printing the Output to output.txt file
# sys.stdout = open('output.txt', 'w')

graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def dfs(v, c):
    if visited[v]:
        if col[v] != c:
            print('-1')
            exit()
        return
    col[v] = c
    visited[v] = 1
    for i in graph[v]:
        dfs(i, c ^ 1)

def bfs(d,v):
    q=[]
    q.append(v)
    visited[v]=1
    while len(q)!=0:
        x=q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i]!=1:
                visited[i]=1
                q.append(i)
        print(x)
    print(l)

def make_graph(e):
    d={}
    for i in range(e):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
        if y not in d.keys():
            d[y] = [x]
        else:
            d[y].append(x)
    return d
def gr2(n):
    d={}
    for i in range(n):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
    return d

def connected_components(graph):
    seen = set()
    def dfs(v):
        vs = set([v])
        component=[]
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component
    ans=[]
    for v in graph:
        if v not in seen:
            d=dfs(v)
            ans.append(d)
    return ans

n,m=mi()
a=li()
b=li()
i=0
j=0
count=0
sa=a[i]
sb=b[j]
while i<n and j<m:
    if sa==sb:
        count+=1
        if i<n-1:
            sa=a[i+1]
        if j<m-1:
            sb=b[j+1]
        i += 1
        j += 1
    elif sa>sb:
        j+=1
        sb+=b[j]
    else:
        i+=1
        sa+=a[i]
print(count)