You are given a tree with N vertices numbered 1, 2, ..., N. The i-th edge connects Vertex a_i and Vertex b_i. You are also given a string S of length N consisting of `0` and `1`. The i-th character of S represents the number of pieces placed on Vertex i.

Snuke will perform the following operation some number of times:

* Choose two pieces the distance between which is at least 2, and bring these pieces closer to each other by 1. More formally, choose two vertices u and v, each with one or more pieces, and consider the shortest path between them. Here the path must contain at least two edges. Then, move one piece from u to its adjacent vertex on the path, and move one piece from v to its adjacent vertex on the path.



By repeating this operation, Snuke wants to have all the pieces on the same vertex. Is this possible? If the answer is yes, also find the minimum number of operations required to achieve it.

Constraints

* 2 \leq N \leq 2000
* |S| = N
* S consists of `0` and `1`, and contains at least one `1`.
* 1 \leq a_i, b_i \leq N(a_i \neq b_i)
* The edges (a_1, b_1), (a_2, b_2), ..., (a_{N - 1}, b_{N - 1}) forms a tree.

Input

Input is given from Standard Input in the following format:


N
S
a_1 b_1
a_2 b_2
:
a_{N - 1} b_{N - 1}


Output

If it is impossible to have all the pieces on the same vertex, print `-1`. If it is possible, print the minimum number of operations required.

Examples

Input

7
0010101
1 2
2 3
1 4
4 5
1 6
6 7


Output

3


Input

7
0010110
1 2
2 3
1 4
4 5
1 6
6 7


Output

-1


Input

2
01
1 2


Output

0
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

"""
・部分木の中で操作を行った場合の、残りdepth和の最小値と最大値を持ってdp
・補助的にコマの個数も持つ
・最大値：ただのdepthの和
・最小値：一番負担の大きな木から他の部分木を引いていった場合だが…
・最小値＋最大値から全部の最大値の和を引く形で更新
"""

N = int(readline())
S = readline().rstrip()
m = map(int,read().split())
AB = zip(m,m)

INF = 10**18

graph = [[] for _ in range(N+1)]
for a,b in AB:
    graph[a].append(b)
    graph[b].append(a)

def solve_rooted(root):
    parent = [0] * (N+1)
    ts = []
    stack = [root]
    while stack:
        x = stack.pop()
        ts.append(x)
        for y in graph[x]:
            if y == parent[x]:
                continue
            parent[y] = x
            stack.append(y)
    max_dep = [0]*(N+1)
    min_dep = [0]*(N+1)
    cnt = [0] * (N+1)
    for i,x in enumerate(S,1):
        cnt[i]=x-ord('0')
    for v in ts[::-1]:
        p = parent[v]
        max_dep_sum = 0
        x = 0
        for c in graph[v]:
            if c==p:
                continue
            cnt[v] += cnt[c]
            d1 = max_dep[c]; d2 = min_dep[c]
            d1 += cnt[c]; d2 += cnt[c]
            max_dep_sum += d1
            if x<d1+d2:
                x=d1+d2
        max_dep[v] = max_dep_sum
        x -= max_dep_sum
        if x<0:
            x=max_dep_sum&1
        min_dep[v] = x
    if min_dep[root]:
        return INF
    return max_dep[root]//2

answer = min(solve_rooted(v) for v in range(1,N+1))
if answer == INF:
    answer = -1
print(answer)