The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.

Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.

Input

The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.

The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.

Output

If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.

Examples

Input

3 3
4 3 5


Output

3


Input

3 4
5 3 4


Output

2


Input

3 7
1 2 3


Output

-1

Note

In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.

In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.

In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1.
# Author: Harrison Whitner
# Date: 02/13/19

# get the first line of input, which contains the number of kegs and size of glass
n, s = list(map(int, input().split()))

# get the next line which contains the keg amounts
kegs = list(map(int, input().split()))

# print -1 if there is not enough kvass
if sum(kegs) < s: print(-1)

# there is enough kvass
else:

    # sort the keg amounts
    kegs.sort(reverse=True)

    i = 0
    while s > 0:

        # find difference between ith keg and last keg (should be least)
        diff = kegs[i] - kegs[n - 1]

        # the ith keg still has more than the least
        if diff > 0:

            # then pull kvass from the ith keg until it is equal to the least
            s -= diff
            kegs[i] -= diff

            # increment i
            if i < n - 2: i += 1

            # TODO: remove this debug pring
            # print('removed', diff, 'liter from the', i, 'th keg')

        # otherwise we have drained the kegs to be less than or equal to the starting least
        else:

            # if all kegs have the same amount and remaining s is greater than n
            if s > n:

                # take an equal amount from all kegs and remove the total from s
                temp = s // n
                s -= temp * n
                kegs = [k - temp for k in kegs]

            # else an equal amount cannot be removed from each keg
            else:

                # take 1 liter from the first keg and subtract 1 liter from the first keg
                # it is not necessary to go through and remove 1 liter from each keg until
                # s is 0, because we know that the least keg amount will be k[0]-1 because
                # s <= n
                kegs[0] -= 1
                s = 0

            # TODO: remove this debug pring
            # print('removed 1 liter from the', i+1, 'th keg')

    # TODO: remove this debug pring
    # print(kegs)

    # print the value of the keg with the least amount
    print(min(kegs))
