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
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define int ll

#define F first
#define S second
#define _sz(x) (int)x.size()

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 1e5 + 20 ;
int n , m , a[N] , b[N] , c[N] , sec[N] ;
vector <int> tmp[N] , g[N] ;

int par[N] ;

inline bool byC (int i , int j) { return c[i] < c[j] ; }

int root (int v) { return par[v] < 0 ? v : par[v] = root(par[v]) ; }

inline void merge (int u , int v) { // c[u] >= c[v]
	if ((u = root(u)) == (v = root(v))) return ;

	g[u].push_back(v) ;
	par[v] = u ;
}

ll sum[N] , res[N] ;

void dfs (int v) {
	ll mx = 0 ;

	sum[v] = b[v] ;

	for (int u : g[v]) {
		dfs(u) ;

		sum[v] += sum[u] ;
		mx = max(mx , sum[u] - max(0ll , res[u] - c[v])) ;
	}

	res[v] = sum[v] + c[v] - mx ;
}

int32_t main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n >> m ;

	for (int i = 0 ; i < n ; i ++) {
		cin >> a[i] >> b[i] ;
		a[i] = max(a[i] , b[i]) ;
		
		c[i] = a[i] - b[i] ;
	}

	iota(sec , sec + n , 0) ;
	sort(sec , sec + n , byC) ;

	for (int i = 0 , u , v ; i < m ; i ++) {
		cin >> u >> v ;
		u -- , v -- ;

		tmp[u].push_back(v) ;
		tmp[v].push_back(u) ;
	}

	for (int i = 0 ; i < n ; i ++) par[i] = -1 ;

	for (int i = 0 ; i < n ; i ++) {
		int v = sec[i] ;
		for (int u : tmp[v])
			if (c[u] <= c[v]) merge(v , u) ;	
	}

	int root = sec[n - 1] ;
	dfs(root) ;

	cout << res[root] << '\n' ;
}
