It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the *i*-th accumulator has *a*<sub class="lower-index">*i*</sub> units of energy. Energy can be transferred from one accumulator to the other. Every time *x* units of energy are transferred (*x* is not necessarily an integer) *k* percent of it is lost. That is, if *x* units were transferred from one accumulator to the other, amount of energy in the first one decreased by *x* units and in other increased by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5da936c833914e398476bcdcd55a9583860c2e5c.png" style="max-width: 100.0%;max-height: 100.0%;"/> units.

Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.
First line of the input contains two integers *n* and *k* (1 ≤ *n* ≤ 10000, 0 ≤ *k* ≤ 99) — number of accumulators and the percent of energy that is lost during transfers.

Next line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ... , *a*<sub class="lower-index">*n*</sub> — amounts of energy in the first, second, .., *n*-th accumulator respectively (0 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 1000, 1 ≤ *i* ≤ *n*).
none
3 50
4 2 1
2.000000000
2 90
1 11
1.909090909
5 26
42 65 23 43 64
45.415178571
5 45
964 515 454 623 594
594.109756098
1 20
784
784.000000000
10 20
812 896 36 596 709 641 679 778 738 302
597.255813953
10 83
689 759 779 927 15 231 976 943 604 917
406.839285714
11 1
235 280 196 670 495 379 391 280 847 875 506
467.586301370
12 71
472 111 924 103 975 527 807 618 400 523 607 424
413.249554367
13 89
19 944 341 846 764 676 222 957 953 481 708 920 950
361.924390244
14 6
256 465 759 589 242 824 638 985 506 128 809 105 301 827
523.427098675
100 95
154 444 715 98 35 347 799 313 40 821 118 786 31 587 888 84 88 751 98 86 321 720 201 247 302 518 663 904 482 385 139 646 581 995 847 775 173 252 508 722 380 922 634 911 102 384 346 212 705 380 220 221 492 421 244 591 758 631 370 866 536 872 294 152 337 810 761 235 789 839 365 366 623 897 905 249 685 838 380 873 702 379 865 68 215 168 425 264 652 228 167 498 733 41 502 21 565 956 430 171
179.075000000
101 71
113 551 568 26 650 547 89 668 64 651 110 515 482 401 170 971 623 672 135 106 985 751 286 255 82 588 122 568 751 867 335 488 324 122 829 256 675 471 255 723 630 802 667 665 206 774 573 499 361 202 620 522 72 220 739 868 101 135 254 519 896 227 224 968 263 826 466 377 360 24 124 874 877 513 130 79 630 786 265 150 232 783 449 914 815 557 646 367 733 576 840 683 417 709 569 432 515 702 811 877 286
343.047284817
102 99
73 348 420 956 955 436 69 714 87 480 102 555 933 215 452 167 157 593 863 816 337 471 371 574 862 967 581 543 330 348 221 640 378 250 500 428 866 379 1 723 880 992 9 419 0 163 800 96 16 25 19 513 653 19 924 144 135 950 449 481 255 582 844 473 189 841 862 520 242 210 573 381 130 820 357 911 884 735 460 428 764 187 344 760 413 636 868 780 123 614 822 869 792 66 636 843 465 449 191 891 819 30
68.702920443
103 26
33 455 273 884 569 636 360 69 802 310 405 594 693 339 43 53 692 514 590 835 1000 191 456 582 641 35 731 207 600 830 416 483 431 377 481 910 367 597 58 413 128 873 42 173 104 553 26 383 673 849 728 503 924 819 108 422 169 454 333 134 926 247 464 289 115 547 567 663 123 396 21 890 385 436 584 432 829 683 345 706 294 901 238 606 12 24 89 882 203 962 804 745 166 425 393 252 415 195 571 596 41 486 445
419.922659430
104 54
683 252 125 813 874 835 651 424 826 139 397 323 143 153 326 941 536 435 317 854 353 222 851 591 420 415 190 872 178 311 612 635 174 505 153 81 559 815 805 414 378 62 75 929 208 942 254 670 329 671 127 494 504 618 292 699 203 959 218 788 285 602 83 104 41 562 272 806 4 582 780 87 639 743 811 263 83 632 230 984 826 304 133 142 612 413 310 985 594 309 787 930 541 92 461 663 675 942 952 610 574 633 758 999
399.430903462
# import sys

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

def read_line():
    return [int(x) for x in input().split()]

def read_int():
    return int(input())

def check(a, mid, k):
    req, take = 0, 0

    for x in a:
        if x > mid:
            take += x-mid
        else:
            req += mid-x

    return take - k*take/100 >= req

def solve():
    _, k = read_line()
    a = read_line()

    low, high, res = 0, 1e9, 0

    for _ in range(100):
        mid = (low + high)/2

        if check(a, mid, k):
            res = low = mid
        else:
            high = mid
    
    print(res)

# t = read_int()
t = 1
while t > 0:
    solve()
    t -= 1