There is a simple undirected graph with N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i connects Vertex U_i and V_i. Also, Vertex i has two predetermined integers A_i and B_i. You will play the following game on this graph.

First, choose one vertex and stand on it, with W yen (the currency of Japan) in your pocket. Here, A_s \leq W must hold, where s is the vertex you choose. Then, perform the following two kinds of operations any number of times in any order:

* Choose one vertex v that is directly connected by an edge to the vertex you are standing on, and move to vertex v. Here, you need to have at least A_v yen in your pocket when you perform this move.
* Donate B_v yen to the vertex v you are standing on. Here, the amount of money in your pocket must not become less than 0 yen.



You win the game when you donate once to every vertex. Find the smallest initial amount of money W that enables you to win the game.

Constraints

* 1 \leq N \leq 10^5
* N-1 \leq M \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* The given graph is connected and simple (there is at most one edge between any pair of vertices).

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_N B_N
U_1 V_1
U_2 V_2
:
U_M V_M


Output

Print the smallest initial amount of money W that enables you to win the game.

Examples

Input

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4


Output

6


Input

5 8
6 4
15 13
15 19
15 1
20 7
1 3
1 4
1 5
2 3
2 4
2 5
3 5
4 5


Output

44


Input

9 10
131 2
98 79
242 32
231 38
382 82
224 22
140 88
209 70
164 64
6 8
1 6
1 4
1 3
4 7
4 9
3 7
3 9
5 9
2 5


Output

582
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

vector<int> par;
int get(int a) {
	if (par[a] < 0) return a;
	return par[a] = get(par[a]);
}

void merge(int a, int b) {
	a = get(a);
	b = get(b);
	assert(a != b);
	par[a] = b;
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0);

	int n, m;
	cin >> n >> m;
	vector<int> a(n), b(n);
	for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
	vector<vector<int>> edges(n);
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		edges[u].push_back(v);
		edges[v].push_back(u);
	}

	par = vector<int>(n, -1);
	vector<int> order(n);
	iota(order.begin(), order.end(), 0);
	sort(order.begin(), order.end(), [&](int i, int j) {
		return a[i] - b[i] < a[j] - b[j];
	});
	vector<ll> bsums(n);
	vector<ll> ans(n);
	vector<bool> done(n, false);
	for (int i = 0; i < n; i++) {
		bsums[i] = b[i];
	}
	for (int i : order) {
		set<int> children;
		for (int j : edges[i]) {
			j = get(j);
			if (done[j]) children.insert(j);
		}
		for (int j : children) {
			bsums[i] += bsums[j];
			merge(j, i);
		}
		if (!children.empty()) {
			ll best = 1e18;
			for (int j : children) {
				best = min(best, bsums[i] - bsums[j] + max<ll>(ans[j], a[i] - b[i]));
			}
			ans[i] = best;
		} else {
			ans[i] = max(a[i], b[i]);
		}
		done[i] = true;
	}
	cout << ans[order.back()] << '\n';

	return 0;
}
