You came to the exhibition and one exhibit has drawn your attention. It consists of n stacks of blocks, where the i-th stack consists of a_i blocks resting on the surface.

The height of the exhibit is equal to m. Consequently, the number of blocks in each stack is less than or equal to m.

There is a camera on the ceiling that sees the top view of the blocks and a camera on the right wall that sees the side view of the blocks.

<image>

Find the maximum number of blocks you can remove such that the views for both the cameras would not change.

Note, that while originally all blocks are stacked on the floor, it is not required for them to stay connected to the floor after some blocks are removed. There is no gravity in the whole exhibition, so no block would fall down, even if the block underneath is removed. It is not allowed to move blocks by hand either.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10^9) — the number of stacks and the height of the exhibit.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ m) — the number of blocks in each stack from left to right.

Output

Print exactly one integer — the maximum number of blocks that can be removed.

Examples

Input

5 6
3 3 3 3 3


Output

10

Input

3 5
1 2 4


Output

3

Input

5 5
2 3 1 4 4


Output

9

Input

1 1000
548


Output

0

Input

3 3
3 1 1


Output

1

Note

The following pictures illustrate the first example and its possible solution.

Blue cells indicate removed blocks. There are 10 blue cells, so the answer is 10.

<image>
#!/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, MutableSequence, defaultdict, deque
# 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, m = map(int, input().split(' '))
    a = sorted(map(int, input().split(' ')), reverse=True)

    m, res = a[0] - 1, a[0] - 1

    for i in range(1, n):
        if a[i] >= m: res, m = res - 1 + a[i], m - 1
        else: res, m = res - m + 2*a[i] - 1, a[i] - 1

    print(res - max(m, 0))


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