Given a tree with n nodes numbered from 1 to n. Each node i has an associated value V_i.

If the simple path from u_1 to u_m consists of m nodes namely u_1 → u_2 → u_3 → ... u_{m-1} → u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = ∑_{i=1}^{m} (-1)^{i+1} ⋅ V_{u_{i}}. A path can also have 0 edges, i.e. u_{1}=u_{m}.

Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo 10^{9}+7. 

Input

The first line contains an integer n (2 ≤ n ≤ 2⋅10^{5} ) — the number of vertices in the tree.

The second line contains n space-separated integers V_1, V_2, …, V_n (-10^9≤ V_i ≤ 10^9) — values of the nodes.

The next n-1 lines each contain two space-separated integers u and v (1≤ u, v≤ n, u ≠ v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.

Output

Print the total sum of alternating functions of all unique simple paths modulo 10^{9}+7. 

Examples

Input

4
-4 1 5 -2
1 2
1 3
1 4


Output

40


Input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8


Output

4

Note

Consider the first example.

A simple path from node 1 to node 2: 1 → 2 has alternating function equal to A(1,2) = 1 ⋅ (-4)+(-1) ⋅ 1 = -5.

A simple path from node 1 to node 3: 1 → 3 has alternating function equal to A(1,3) = 1 ⋅ (-4)+(-1) ⋅ 5 = -9.

A simple path from node 2 to node 4: 2 → 1 → 4 has alternating function A(2,4) = 1 ⋅ (1)+(-1) ⋅ (-4)+1 ⋅ (-2) = 3.

A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 ⋅ (-4) = -4.

Similarly, A(2, 1) = 5, A(3, 1) = 9, A(4, 2) = 3, A(1, 4) = -2, A(4, 1) = 2, A(2, 2) = 1, A(3, 3) = 5, A(4, 4) = -2, A(3, 4) = 7, A(4, 3) = 7, A(2, 3) = 10, A(3, 2) = 10. So the answer is (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40.

Similarly A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 which sums upto 40. 
import sys
range = xrange
input = raw_input 
MOD = 10**9 + 7

def exclusive(A, base = 0, f = lambda a,b: a + b):
    n = len(A)
    B = [base] * n
    for bit in reversed(range(n.bit_length())):
        for i in reversed(range(n)):
            B[i] = B[i // 2]
        for i in range(n & ~(bit == 0)):
            ind = (i >> bit) ^ 1
            B[ind] = f(B[ind], A[i])
    return B

def rerooter(graph, A, f = lambda a,b: a + b):
    n = len(graph)

    P = [0] * n
    bfs = [0]
    for node in bfs:
        for nei in graph[node]:
            if P[node] != nei:
                P[nei] = node
                bfs.append(nei)

    DP = list(A)
    for node in reversed(bfs[1:]):
        DP[P[node]] = f(DP[P[node]], DP[node])
    
    outDP = [[]] * n
    inDP = [[]] * n
    for node in bfs:
        DP[P[node]] = DP[node]
        outDP[node] = DPvec = [DP[nei] for nei in graph[node]]
        inDP[node] = DPexcl_vec = exclusive(DPvec, A[node], f)

        for nei, dp in zip(graph[node], DPexcl_vec):
            DP[nei] = dp
    
    rootDP = [f(inDP[node][0], outDP[node][0]) for node in range(n)]
    return rootDP, outDP, inDP

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
 
n = inp[ii]; ii += 1
V = inp[ii: ii + n]; ii += n
 
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
    u = inp[ii] - 1; ii += 1
    v = inp[ii] - 1; ii += 1
    coupl[u].append(v)
    coupl[v].append(u)
 
def adder((a1,b1), (a2,b2)):
    return a1 + b2, b1 + a2
DP, outDP, inDP = rerooter(coupl, [(1,0)]*n, adder)
 
s = sum(DP[node][0] * V[node] for node in range(n))
for node in range(n):
    for nei, (a1,b1), (a2,b2) in zip(coupl[node], outDP[node], inDP[node]):
        if node < nei:
            s += (a1 * b2 - a2 * b1) * (V[nei] - V[node])
print s % MOD
