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;
const int maxn = 100100;
int a[maxn], b[maxn], c[maxn], uf[maxn], act[maxn], od[maxn];
vector<int> to[maxn];
long long sum[maxn], f[maxn];
int find(int x){
	if(x == uf[x]) return x;
	return uf[x] = find(uf[x]);
}
int main(){
	iota(uf, uf + maxn, 0);
	iota(od, od + maxn, 0);
	int n, m; cin >> n >> m;
	for(int i = 0; i < n; i++) cin >> a[i] >> b[i];
	for(int i = 0; i < m; i++){
		int u, v; cin >> u >> v; u--; v--;
		to[u].push_back(v);
		to[v].push_back(u);
	}
	for(int i = 0; i < n; i++) c[i] = max(a[i] - b[i], 0), sum[i] = b[i];
	sort(od, od + n, [&](int a, int b){
		return c[a] < c[b];
	});
	for(int i = 0; i < n; i++){
		int u = od[i];
		vector<int> comp;
		for(int v : to[u]){
			if(act[find(v)]){
				comp.push_back(find(v));
			}
		}
		sort(comp.begin(), comp.end());
		comp.erase(unique(comp.begin(), comp.end()), comp.end());
		for(int v : comp){
			sum[u] += sum[v];
		}
		if(comp.size()){
			f[u] = 1ll<<60;
			for(int v : comp){
				f[u] = min(f[u], sum[u] - sum[v] + max(1ll * c[u], f[v]));
			}
		} else {
			f[u] = c[u] + b[u];
		}
		for(int v : comp)
			uf[v] = u;
		act[u] = 1;
	}
	cout << f[od[n-1]] << endl;
	return 0;
}
