There are n districts in the town, the i-th district belongs to the a_i-th bandit gang. Initially, no districts are connected to each other.

You are the mayor of the city and want to build n-1 two-way roads to connect all districts (two districts can be connected directly or through other connected districts).

If two districts belonging to the same gang are connected directly with a road, this gang will revolt.

You don't want this so your task is to build n-1 two-way roads in such a way that all districts are reachable from each other (possibly, using intermediate districts) and each pair of directly connected districts belong to different gangs, or determine that it is impossible to build n-1 roads to satisfy all the conditions.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 500) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (2 ≤ n ≤ 5000) — the number of districts. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), where a_i is the gang the i-th district belongs to.

It is guaranteed that the sum of n does not exceed 5000 (∑ n ≤ 5000).

Output

For each test case, print:

  * NO on the only line if it is impossible to connect all districts satisfying the conditions from the problem statement. 
  * YES on the first line and n-1 roads on the next n-1 lines. Each road should be presented as a pair of integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i), where x_i and y_i are two districts the i-th road connects. 



For each road i, the condition a[x_i] ≠ a[y_i] should be satisfied. Also, all districts should be reachable from each other (possibly, using intermediate districts).

Example

Input


4
5
1 2 2 1 3
3
1 1 1
4
1 1000 101 1000
4
1 2 3 4


Output


YES
1 3
3 5
5 4
1 2
NO
YES
1 2
2 3
3 4
YES
1 2
1 3
1 4
# _______________________________________________________________#
from math import *
import threading
import sys


def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x - 1)


def lower_bound(li, num):
	answer = -1
	start = 0
	end = len(li) - 1

	while (start <= end):
		middle = (end + start) // 2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer  # index where x is not less than num


def upper_bound(li, num):
	answer = -1
	start = 0
	end = len(li) - 1

	while (start <= end):
		middle = (end + start) // 2

		if li[middle] <= num:
			answer = middle
			start = middle + 1

		else:
			end = middle - 1
	return answer  # index where x is not greater than num


def abs(x):
	return x if x >= 0 else -x


def binary_search(li, val, lb, ub):
	ans = -1
	while (lb <= ub):
		mid = (lb + ub) // 2
		# print(mid, li[mid])
		if li[mid] > val:
			ub = mid - 1
		elif val > li[mid]:	
			lb = mid + 1
		else:
			ans = mid # return index
			break
	return ans


def kadane(x):  # maximum sum contiguous subarray
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = max(sum_so_far, current_sum)
	return sum_so_far


def pref(li):
	pref_sum = [0]
	for i in li:
		pref_sum.append(pref_sum[-1] + i)
	return pref_sum


def graph(n, m):
	adj = dict()
	for i in range(1, n + 1):
		adj.setdefault(i, 0)
	for i in range(m):
		a, b = map(int, input().split())
		adj[a] += 1
		adj[b] += 1
	return adj


def SieveOfEratosthenes(n):
	prime = [True for i in range(n+1)]
	p = 2
	li = []
	while (p * p <= n):
		if (prime[p] == True):
			for i in range(p * p, n+1, p):
				prime[i] = False
		p += 1

	for p in range(2, len(prime)):
		if prime[p]:
			li.append(p)
	return li


# _______________________________________________________________#


INF = 1e9
# sys.setrecursionlimit(900000)
# threading.stack_size(10**5)  # remember it cause mle
# def main():
primes = SieveOfEratosthenes(1300000)
def fn(n):
	i = 2
	res = n-1
	val = 0
	for i in range(int(sqrt(n))+1,1,-1):
	
		val = i - 1 + int(ceil((n-i)/i))
		#print(val,res)
		if val <= res:
			#print(i)
			res = val
		else:
			break
		
	return res


def solve(a,b,c): 
	dis = (b**2) - (4 * a*c) 

	# find two results 
	ans1 = (-b-sqrt(dis))/(2 * a) 
	#ans2 = (-b + cmath.sqrt(dis))/(2 * a) 

	# printing the results 
	#print('The roots are') 
	print(int(ans1)) 
	#print(ans2) 	
	return
	

for _ in range(int(input())):
#for _ in range(1):
	n = int(input())
	#n,m = map(int, input().split())
	# s = list(input())
	# a = [int(x) for x in s]
	# s = list(input())
	a = list(map(int, input().split()))
	#b = list(map(int,input().split()))
	#c = list(map(int,input().split()))
	# adj = graph(n,m)
	if len(set(a)) == 1:
		print('NO')
	else:
		print('YES')
		one, two = list(set(a))[:2]
		ind_one, ind_two = a.index(one), a.index(two)
		for i in range(1, n):
			if a[i] == one and i != ind_two:
				print(ind_two+1, i+1)
			else:
				print(ind_one+1, i+1)


			





















'''		
t = threading.Thread(target=main)
t.start()
t.join()
'''
