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;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
const int MAXN = 100005;
#define sz(v) ((int)v.size())

struct disj{
	int pa[MAXN];
	void init(int n){
		iota(pa, pa + n + 1, 0);
	}
	int find(int x){
		return pa[x] = (pa[x] == x ? x : find(pa[x]));
	}
	bool uni(int p, int q){
		p = find(p);
		q = find(q);
		if(p == q) return 0;
		pa[q] = p; return 1;
	}
}disj;

struct func{
	lint out;
	lint B, A; // f(W) = max(W + B, A)
}val[MAXN];

vector<int> gph[MAXN];
int n, m, a[MAXN], b[MAXN];
int rev[MAXN];

int main(){
	scanf("%d %d",&n,&m);
	for(int i=1; i<=n; i++){
		scanf("%d %d",&a[i],&b[i]);
		a[i] = max(a[i], b[i]);
	}
	vector<int> ord(n);
	iota(ord.begin(), ord.end(), 1);
	sort(ord.begin(), ord.end(), [&](const int &x, const int &y){
		return a[x] - b[x] < a[y] - b[y];
	});
	for(int i=0; i<sz(ord); i++) rev[ord[i]] = i;
	for(int i=0; i<m; i++){
		int s, e;
		scanf("%d %d",&s,&e);
		s = rev[s]; e = rev[e];
		if(s > e) swap(s, e);
		gph[ord[e]].push_back(ord[s]);
	}
	disj.init(n);
	for(auto &x : ord){
		vector<func> vect;
		for(auto &i : gph[x]){
			int p = disj.find(i);
			if(disj.find(x) == p) continue;
			vect.push_back(val[p]);
			disj.uni(i, x);
		}
		lint bsum = 0;
		for(auto &i : vect) bsum += i.B;
		func ret = {(lint)1e18, b[x] + bsum, a[x] + bsum};
		for(auto &i : vect){
			lint cur_out = max(i.out + b[x], 1ll * a[x]);
			cur_out += bsum - i.B;
			ret.out = min(ret.out, cur_out);
		}
		if(vect.empty()) ret.out = a[x];
		val[disj.find(x)] = ret;
	}
	cout << val[disj.find(ord.back())].out << endl;
}
