Parsa has a humongous tree on n vertices.

On each vertex v he has written two integers l_v and r_v.

To make Parsa's tree look even more majestic, Nima wants to assign a number a_v (l_v ≤ a_v ≤ r_v) to each vertex v such that the beauty of Parsa's tree is maximized.

Nima's sense of the beauty is rather bizarre. He defines the beauty of the tree as the sum of |a_u - a_v| over all edges (u, v) of the tree.

Since Parsa's tree is too large, Nima can't maximize its beauty on his own. Your task is to find the maximum possible beauty for Parsa's tree.

Input

The first line contains an integer t (1≤ t≤ 250) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (2≤ n≤ 10^5) — the number of vertices in Parsa's tree.

The i-th of the following n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9).

Each of the next n-1 lines contains two integers u and v (1 ≤ u , v ≤ n, u≠ v) meaning that there is an edge between the vertices u and v in Parsa's tree.

It is guaranteed that the given graph is a tree.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case print the maximum possible beauty for Parsa's tree.

Example

Input


3
2
1 6
3 8
1 2
3
1 3
4 6
7 9
1 2
2 3
6
3 14
12 20
12 19
2 12
10 17
3 17
3 2
6 5
1 5
2 6
4 6


Output


7
8
62

Note

The trees in the example:

<image>

In the first test case, one possible assignment is a = \{1, 8\} which results in |1 - 8| = 7.

In the second test case, one of the possible assignments is a = \{1, 5, 9\} which results in a beauty of |1 - 5| + |5 - 9| = 8
import io
import os

from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from heapq import heappush, heappop, heapify
from math import gcd, inf

from types import GeneratorType

# From https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/bootstrap.py
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc


def solve(N, LR, edges):

    graph = [[] for i in range(N)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    dp = [-1] * (2 * N)

    @bootstrap
    def dfs(u, p, i):
        if dp[2 * u + i] == -1:
            # i is 0 or 1
            a = LR[u][i]
            best = 0.0
            for v in graph[u]:
                if v != p:
                    yield dfs(v, u, 0)
                    yield dfs(v, u, 1)
                    best += max(
                        abs(a - LR[v][0]) + dp[2 * v + 0],
                        abs(a - LR[v][1]) + dp[2 * v + 1],
                    )
            dp[2 * u + i] = best
        yield

    root = 0
    dfs(root, -1, 0)
    dfs(root, -1, 1)
    return int(max(dp[2 * root + 0], dp[2 * root + 1]))


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = int(input())
    for tc in range(1, TC + 1):
        (N,) = [int(x) for x in input().split()]

        LR = [[int(x) for x in input().split()] for i in range(N)]
        edges = [
            [int(x) - 1 for x in input().split()] for i in range(N - 1)
        ]  # 0 indexed
        ans = solve(N, LR, edges)
        print(ans)
