Problem: You are given k sequences of integers. The length of the i-th sequence equals to n_i.

You have to choose exactly two sequences i and j (i ≠ j) such that you can remove exactly one element in each of them in such a way that the sum of the changed sequence i (its length will be equal to n_i - 1) equals to the sum of the changed sequence j (its length will be equal to n_j - 1).

Note that it's required to remove exactly one element in each of the two chosen sequences.

Assume that the sum of the empty (of the length equals 0) sequence is 0.

Input

The first line contains an integer k (2 ≤ k ≤ 2 ⋅ 10^5) — the number of sequences.

Then k pairs of lines follow, each pair containing a sequence.

The first line in the i-th pair contains one integer n_i (1 ≤ n_i < 2 ⋅ 10^5) — the length of the i-th sequence. The second line of the i-th pair contains a sequence of n_i integers a_{i, 1}, a_{i, 2}, ..., a_{i, n_i}.

The elements of sequences are integer numbers from -10^4 to 10^4.

The sum of lengths of all given sequences don't exceed 2 ⋅ 10^5, i.e. n_1 + n_2 + ... + n_k ≤ 2 ⋅ 10^5.

Output

If it is impossible to choose two sequences such that they satisfy given conditions, print "NO" (without quotes). Otherwise in the first line print "YES" (without quotes), in the second line — two integers i, x (1 ≤ i ≤ k, 1 ≤ x ≤ n_i), in the third line — two integers j, y (1 ≤ j ≤ k, 1 ≤ y ≤ n_j). It means that the sum of the elements of the i-th sequence without the element with index x equals to the sum of the elements of the j-th sequence without the element with index y.

Two chosen sequences must be distinct, i.e. i ≠ j. You can print them in any order.

If there are multiple possible answers, print any of them.

Examples

Input

2
5
2 3 1 3 2
6
1 1 2 2 2 1


Output

YES
2 6
1 2


Input

3
1
5
5
1 1 1 1 1
2
2 3


Output

NO


Input

4
6
2 2 2 2 2 2
5
2 2 2 2 2
3
2 2 2
5
2 2 2 2 2


Output

YES
2 2
4 1

Note

In the first example there are two sequences [2, 3, 1, 3, 2] and [1, 1, 2, 2, 2, 1]. You can remove the second element from the first sequence to get [2, 1, 3, 2] and you can remove the sixth element from the second sequence to get [1, 1, 2, 2, 2]. The sums of the both resulting sequences equal to 8, i.e. the sums are equal.

Solution (in PYTHON3):

Solution:
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############

"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
 
import os,sys
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
 
 
def ii():  return int(input())
def si():  return input()
def mi():  return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li():  return list(mi())
 
def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
    
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list  m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scanner = lambda: int(input())
string = lambda: input().rstrip()
get_list = lambda: list(read())
read = lambda: map(int, input().split())
get_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;

 
def ceil(x):
    if x != int(x):
        x = int(x) + 1
    return x
 
def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val

def fact(x):
	val = 1
	while x > 0:
		val *= x
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## lcm function
def lcm(a, b):
	return (a * b) // math.gcd(a, b)

def is_integer(n):
	return math.ceil(n) == math.floor(n)
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
	if k > n:
		return 0
	if(k > n - k):
		k = n - k
	res = 1
	for i in range(k):
		res = res * (n - i)
		res = res / (i + 1)
	return int(res)
 
## upper bound function code -- such that e in a[:i] e < x;

 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    prime[0], prime[1] = False, False
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

# Euler's Toitent Function phi
def phi(n) : 
  
    result = n 
    p = 2
    while(p * p<= n) : 
        if (n % p == 0) : 
            while (n % p == 0) : 
                n = n // p 
            result = result * (1.0 - (1.0 / (float) (p))) 
        p = p + 1
    if (n > 1) : 
        result = result * (1.0 - (1.0 / (float)(n))) 
   
    return (int)(result) 

def is_prime(n):
	if n == 0:
		return False
	if n == 1:
		return True
	for i in range(2, int(n ** (1 / 2)) + 1):
		if not n % i:
			return False
 
	return True

def next_prime(n, primes):
	while primes[n] != True:
		n += 1
	return n

 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    res = []
    for i in range(2, int(x ** 0.5) + 1):
    	while x % i == 0:
    		res.append(i)
    		x //= i
    if x != 1:
   		res.append(x)
    return res
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

def factors(n):
	res = []
	for i in range(1, int(n ** 0.5) + 1):
		if n % i == 0:
			res.append(i)
			res.append(n // i)
	return list(set(res))
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();

def binary_search(low, high, w, h, n):
	while low < high:
		mid = low + (high - low) // 2
		# print(low, mid, high)
		if check(mid, w, h, n):
			low = mid + 1
		else:
			high = mid
	return low

## for checking any conditions
def check(beauty, s, n, count):
	pass
	

 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
alphs = "abcdefghijklmnopqrstuvwxyz"
 
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
 
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
# import numpy as np

def solve():
	k = scanner()
	trippies = []
	for i in range(k):
		n = scanner()
		sequence = get_list()
		summ = sum(sequence)
		for j, s in enumerate(sequence):
			trippies.append([summ - s, i + 1, j + 1])
	trippies.sort()
	for i in range(1, len(trippies)):
		if trippies[i][0] == trippies[i - 1][0] and trippies[i][1] != trippies[i - 1][1]:
			print("YES")
			print(trippies[i][1], trippies[i][2])
			print(trippies[i - 1][1], trippies[i - 1][2])
			return
	print("NO")
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
 
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)
 
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
# endregion
 
 
if __name__ == "__main__":
    #read()
    # sys.stdin = open("input.txt", "r")
    # sys.stdout = open("output.txt", "w")
    for i in range(1):
    	solve()
    #dmain()
 
# Comment Read()
	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))