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

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)

found = [0]*n
children = [[] for _ in range(n)]
P = [-1]*n

bfs = [0]
P[0] = 0
for node in bfs:
    for nei in coupl[node]:
        if P[nei] == -1:
            bfs.append(nei)
            P[nei] = node
            children[node].append(nei)


E = [0]*n
O = [0]*n

for node in reversed(bfs):
    O[node] += 1
    p = P[node]
    if p != node:
        E[p] += O[node]
        O[p] += E[node]

UE = list(E)
UO = list(O)

for node in bfs:
    for child in children[node]:
        UO[child] += UE[node] - O[child]
        UE[child] += UO[node] - E[child]

out = []
for node in range(n):

    ue = UE[node]
    count = uo = UO[node]
    p = P[node]
    if p != node:
        e = UE[p] - O[node]
        o = UO[p] - E[node]

        count += - E[node] * o + O[node] * e
    for child in children[node]:
        e = ue - O[child]
        o = uo - E[child]

        count += E[child] * o - O[child] * e
    out.append(count)
print sum(out[i] * V[i] for i in range(n)) % (10**9 + 7)
