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.
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.

Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <hello@cheran.io>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
from collections import Counter
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush

if sys.version_info[0] < 3:
    # from cPickle import dumps
    from io import BytesIO as stream
    # from Queue import PriorityQueue, Queue
else:
    # from functools import reduce
    from io import StringIO as stream
    from math import gcd
    # from pickle import dumps
    # from queue import PriorityQueue, Queue


if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    def gcd(x, y):
        """gcd(x, y) -> int
        greatest common divisor of x and y
        """
        while y:
            x, y = y, x % y
        return x

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip


def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.

    Args:
        sync (bool, optional): The new synchronization setting.

    """
    global input, flush

    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')

        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))


def main():
    n = int(input())
    a = list(map(float, input().split(' ')))

    total_sum = sum(a)
    duplicate_cnt = Counter(a)

    res = []
    for i, ai in enumerate(a):
        nice_val = (total_sum - ai) / 2
        if nice_val == ai:
            if duplicate_cnt[nice_val] > 1:
                res.append(i + 1)
        else:
            if nice_val in duplicate_cnt:
                res.append(i + 1)

    print(len(res))
    print(*res)


if __name__ == '__main__':
    sync_with_stdio(False)

    if 'PyPy' in sys.version:
        from _continuation import continulet

        def bootstrap(c):
            callable, arg = c.switch()
            while True:
                to = continulet(lambda _, f, x: f(x), callable, arg)
                callable, arg = c.switch(to=to)

        c = continulet(bootstrap)
        c.switch()

        main()

    else:
        import threading

        sys.setrecursionlimit(2097152)
        threading.stack_size(134217728)

        main_thread = threading.Thread(target=main)
        main_thread.start()
        main_thread.join()
