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
#include <bits/stdc++.h>
using namespace std;

#ifdef SG
	#include <debug.h>
#else
	template<typename T> struct outputer;
	struct outputable {};
	#define PRINT(...)
	#define OUTPUT(...)
	#define show(...)
	#define debug(...)
	#define deepen(...)
	#define timer(...)
	#define fbegin(...)
	#define fend
	#define pbegin(...)
	#define pend
#endif

#define ARG4(_1,_2,_3,_4,...) _4

#define forn3(i,l,r) for (int i = int(l); i < int(r); ++i)
#define forn2(i,n) forn3 (i, 0, n)
#define forn(...) ARG4(__VA_ARGS__, forn3, forn2) (__VA_ARGS__)

#define ford3(i,l,r) for (int i = int(r) - 1; i >= int(l); --i)
#define ford2(i,n) ford3 (i, 0, n)
#define ford(...) ARG4(__VA_ARGS__, ford3, ford2) (__VA_ARGS__)

#define ve vector
#define pa pair
#define tu tuple
#define mp make_pair
#define mt make_tuple
#define pb emplace_back
#define fs first
#define sc second
#define all(a) (a).begin(), (a).end()
#define sz(a) ((int)(a).size())

typedef long double ld;
typedef int64_t ll;
typedef uint64_t ull;
typedef uint32_t ui;
typedef uint16_t us;
typedef uint8_t uc;
typedef pa<int, int> pii;
typedef pa<int, ll> pil;
typedef pa<ll, int> pli;
typedef pa<ll, ll> pll;
typedef ve<int> vi;

template<typename T> inline auto sqr (T x) -> decltype(x * x) {return x * x;}
template<typename T1, typename T2> inline bool umx (T1& a, T2 b) {if (a < b) {a = b; return 1;} return 0;}
template<typename T1, typename T2> inline bool umn (T1& a, T2 b) {if (b < a) {a = b; return 1;} return 0;}

const int N = 100000;

struct Input {
	int n;
	vi e[N];
	int l[N], r[N];
	
	bool read() {
		if (!(cin >> n)) {
			return 0;
		}
		forn (i, n) {
			scanf("%d%d", &l[i], &r[i]);
		}
		forn (i, n - 1) {
			int x, y;
			scanf("%d%d", &x, &y);
			--x;
			--y;
			e[x].pb(y);
			e[y].pb(x);
		}
		return 1;
	}

	void init(const Input &input) {
		*this = input;
	}
};

struct Data: Input {
	ll ans;

	void write() {
		cout << ans << endl;	
	}
};


namespace Main {
	
	struct Solution: Data {
		ll d[N][2];

		void calc(int v, int pr) {
			d[v][0] = 0;
			d[v][1] = 0;
			for (int to : e[v]) {
				if (to == pr) {
					continue;
				}
				calc(to, v);
				forn (t1, 2) {
					int v1 = (t1 == 0 ? l[v] : r[v]);
					ll mx = 0;
					forn (t2, 2) {
						int v2 = (t2 == 0 ? l[to] : r[to]);
						umx(mx, abs(v1 - v2) + d[to][t2]);
					}
					d[v][t1] += mx;
				}
			}
		}
		
		void solve() {
			debug(e, e + n);
			debug(l, l + n);
			debug(r, r + n);
			calc(0, -1);
			forn (i, n) {
				debug(d[i], d[i] + 2);
			}
			ans = max(d[0][0], d[0][1]);
		}
		
		void clear() {
			*this = Solution();
		}
	};
}


Main::Solution sol;

int main() {
	cout.setf(ios::showpoint | ios::fixed);
	cout.precision(20);

	#ifdef SG
		freopen((problemname + ".in").c_str(), "r", stdin);
//		freopen((problemname + ".out").c_str(), "w", stdout);
	#endif
	
	int t;
	cin >> t;
	forn (i, t) {
		sol.read();
		sol.solve();
		sol.write();
		sol.clear();
	}
	
	return 0;
}
