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
#ifdef _MSC_VER
#define _CRT_SECURE_NO_DEPRECATE
#pragma comment(linker, "/STACK:66777216")
#include <cstdio>
#define GETS gets_s
#else
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#include <stdio.h>
#define GETS gets
#endif
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iostream>
#include <iterator>
#include <cmath>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <ctime>
#include <cstring>
#include <functional>
#include <chrono>
using namespace std;
#define pb push_back
#define ppb pop_back
#define pi 3.1415926535897932384626433832795028841971
#define mp make_pair
#define x first
#define y second
#define pii pair<int,int>
#define pdd pair<double,double>
#define INF 1000000000
#define FOR(i,a,b) for (int _n(b), i(a); i <= _n; i++)
#define FORD(i,a,b) for(int i=(a),_b=(b);i>=_b;i--)
#define all(c) (c).begin(), (c).end()
#define SORT(c) sort(all(c))
#define rep(i,n) FOR(i,1,(n))
#define rept(i,n) FOR(i,0,(n)-1)
#define L(s) (int)((s).size())
#define C(a) memset((a),0,sizeof(a))
#define VI vector <int>
#define ll long long

int a, b, c, d, n, m, k, kolt;
VI sm[100002];
pii mas[100002];
ll mem[100002][2];

ll rec(int v, int pr, int cc) {
	if (mem[v][cc] != -1) return mem[v][cc];

	int cv = cc ? mas[v].y : mas[v].x;
	ll cur = 0;
	rept(i, L(sm[v])) {
		int w = sm[v][i];
		if (w == pr) continue;
		ll c1 = rec(w, v, 0) + abs(cv - mas[w].x);
		ll c2 = rec(w, v, 1) + abs(cv - mas[w].y);
		cur += max(c1, c2);
	}

	return mem[v][cc] = cur;
}

ll solve() {
	rept(i, n) mem[i][0] = mem[i][1] = -1;

	ll a1 = rec(0, -1, 0);
	ll a2 = rec(0, -1, 1);

	return max(a1, a2);
}
int main() {
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);

	scanf("%d", &kolt);
	rep(hod, kolt) {
		scanf("%d", &n);
		rept(i, n) {
			scanf("%d%d", &mas[i].x, &mas[i].y);
		}
		rept(i, n) sm[i].clear();
		rept(i, n - 1) {
			scanf("%d%d", &a, &b); --a; --b;
			sm[a].push_back(b); sm[b].push_back(a);
		}

		ll ans = solve();
		cout << ans << "\n";
	}
}
