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;
vector <int> g[100010], t[100010];
int a[100010], b[100010];
int par[100010];
long long cost[100010];
long long sub[100010];

int main(int argc, char const *argv[])
{
	int n, m;
	cin >> n >> m;
	vector <int> ord;
	for(int i = 1; i <= n; i++) {
		cin >> a[i] >> b[i];
		par[i] = i;
		ord.push_back(i);
	}
	for(int i = 1; i <= m; i++) {
		int x, y; cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	sort(ord.begin(), ord.end(), [&] (int i, int j) { return a[i] - b[i] < a[j] - b[j]; });
	
	vector <int> done (n + 1, false);
	function <int(int)>  root = [&] (int x) {
		if(par[x] == x) return x;
		else return par[x] = root(par[x]);
	};
	for(int i : ord) {
		for(auto j : g[i]) {
			if(done[j] && root(j) != i) {
				t[i].push_back(root(j));
				par[root(j)] = i;
			}
		}
		done[i] = true;
	}
	function <void(int)> dfs = [&] (int x) {
		sub[x] = b[x];
		for(auto i : t[x]) {
			dfs(i);
			sub[x] += sub[i];
		}
		long long var = max(0, a[x] - b[x]);
		cost[x] = sub[x] + var;
		for(auto i : t[x]) {			
			cost[x] = min(cost[x], sub[x] - sub[i] + max(var, cost[i]));
		}
	};
	dfs(root(1));
	cout << cost[root(1)] << endl;
	return 0;
}