Let's call an array good if there is an element in the array that equals to the sum of all other elements. For example, the array a=[1, 3, 3, 7] is good because there is the element a_4=7 which equals to the sum 1 + 3 + 3.

You are given an array a consisting of n integers. Your task is to print all indices j of this array such that after removing the j-th element from the array it will be good (let's call such indices nice).

For example, if a=[8, 3, 5, 2], the nice indices are 1 and 4: 

  * if you remove a_1, the array will look like [3, 5, 2] and it is good; 
  * if you remove a_4, the array will look like [8, 3, 5] and it is good. 



You have to consider all removals independently, i. e. remove the element, check if the resulting array is good, and return the element into the array.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — elements of the array a.

Output

In the first line print one integer k — the number of indices j of the array a such that after removing the j-th element from the array it will be good (i.e. print the number of the nice indices).

In the second line print k distinct integers j_1, j_2, ..., j_k in any order — nice indices of the array a.

If there are no such indices in the array a, just print 0 in the first line and leave the second line empty or do not print it at all.

Examples

Input


5
2 5 1 2 2


Output


3
4 1 5

Input


4
8 3 5 2


Output


2
1 4 


Input


5
2 1 2 4 3


Output


0

Note

In the first example you can remove any element with the value 2 so the array will look like [5, 1, 2, 2]. The sum of this array is 10 and there is an element equals to the sum of remaining elements (5 = 1 + 2 + 2).

In the second example you can remove 8 so the array will look like [3, 5, 2]. The sum of this array is 10 and there is an element equals to the sum of remaining elements (5 = 3 + 2). You can also remove 2 so the array will look like [8, 3, 5]. The sum of this array is 16 and there is an element equals to the sum of remaining elements (8 = 3 + 5).

In the third example you cannot make the given array good by removing exactly one element.
## necessary imports
import sys
input = sys.stdin.readline
from math import ceil, floor, factorial;

# 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 a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    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;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 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):
            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)] 
    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

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 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):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

n = int(input()); a = int_array();
ans = []; s = sum(a);
for i in range(n):
    a[i] = (a[i], i);
a.sort();
for i in range(n-1, -1, -1):
    k = s - a[i][0];
    if i == n - 1:
        k -= a[n-2][0];
        if k == a[n-2][0]:
            ans.append(a[i][1]+1);
    else:
        k -= a[-1][0];
        if k == a[-1][0]:
            ans.append(a[i][1] + 1);
print(len(ans));
print(*ans);