Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes. 

Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.

Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.

Input

The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.

The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.

Output

Print a single integer — the answer to the problem modulo mod.

Examples

Input

3 3 3 100
1 1 1


Output

10


Input

3 6 5 1000000007
1 2 3


Output

0


Input

3 5 6 11
1 2 1


Output

0
"""
________        _____________              ______
___  __ \____  ____  __ \__(_)__   _______ ___  /
__  /_/ /_  / / /_  /_/ /_  /__ | / /  __ `/_  /
_  ____/_  /_/ /_  _, _/_  / __ |/ // /_/ /_  /
/_/     _\__, / /_/ |_| /_/  _____/ \__,_/ /_/
        /____/

https://github.com/Cheran-Senthil/PyRival
Copyright (c) 2018 Cheran Senthilkumar
"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
import sys
from atexit import register
from bisect import bisect_left, bisect_right

# import random
# from collections import Counter, MutableSequence, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from heapq import heappop, heappush

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


if sys.version_info[0] < 3:
    class dict(dict):
        def items(self):
            return dict.iteritems(self)

        def keys(self):
            return dict.iterkeys(self)

        def values(self):
            return dict.itervalues(self)

    input = raw_input
    range = xrange

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


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

    Parameters
    ----------
    sync : bool, optional
        The new synchronization setting. Default is True.
    """
    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, b, mod = map(int, input().split(' '))
    a = list(map(int, input().split(' ')))
    dp = [[0] * (b + 1) if i > 0 else [1] * (b + 1) for i in range(m + 1)]

    for i in range(n):
        for j in range(1, m + 1):
            for k in range(a[i], b + 1):
                dp[j][k] = (dp[j - 1][k - a[i]] + dp[j][k]) % mod

    print(dp[m][b])


if __name__ == '__main__':
    sys.setrecursionlimit(10000)
    sync_with_stdio(False)
    main()
