Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. 

Iahub asks Iahubina: can you build a rooted tree, such that
 -  each internal node (a node with at least one son) has at least two sons; -  node *i* has *c*<sub class="lower-index">*i*</sub> nodes in its subtree? 
Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain *n* nodes.
The first line of the input contains integer *n* (1 ≤ *n* ≤ 24). Next line contains *n* positive integers: the *i*-th number represents *c*<sub class="lower-index">*i*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ *n*).
#!/usr/bin/python

import sys

def Ni(): return tuple(map(int, sys.stdin.readline().split()))

n = Ni()[0]
c = Ni()

freq = [c.count(i) for i in range(n+1)]

answer = "NO"

def backtrack(stack, avail, sumleft):
    #print stack, avail, sumleft
    if len(stack) == 1 and sumleft == 0:
        print "YES"
        sys.exit(0)

    # try to shift a 1
    if avail[1] > 0:
        avail[1] -= 1
        backtrack(stack + [1], avail, sumleft - 1)
        avail[1] += 1

    # reduce if possible
    if len(stack) < 2:
        return

    s = 1 + stack[-1]
    for i in range(2, len(stack) + 1):
        s += stack[-i]
        if s >= n + 1:
            break

        if avail[s] > 0:
            avail[s] -= 1
            backtrack(stack[:-i] + [s], avail, sumleft - s)
            avail[s] += 1


backtrack([], freq, sum(c))
print "NO"
