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
//#define __USE_MINGW_ANSI_STDIO 0
#include <iostream>
#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout<<#x<<" :: "<<x<<endl;
#define debug2(x,y) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<endl;
#define debug3(x,y,z) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<endl;

#define boost ios::sync_with_stdio(0); cin.tie(0)

#define fi first
#define se second

typedef long long ll;
typedef double ld;
typedef pair<int, int> pii;

const int N = 1e6 + 5;

/***************************************************************************/

int A[N], B[N];
vector<int> adj[N];
vector<pii> ord;

int par[N];
ll summ[N], DP[N];

int find(int i) {
	if(par[i] != i) par[i] = find(par[i]);
	return par[i];
} 

void Union(int a, int b) {
	a = find(a), b = find(b);
	if(a == b) return;
	par[b] = a;
}

set<int> S;
int done[N];

int main() {

	boost;
	int n, m; cin>>n>>m;
	for(int i=1; i<=n; i++) {
		cin>>A[i]>>B[i];
		ord.push_back(pii(max(A[i]-B[i], 0), i));
		par[i] = i;
	}
	for(int i=1; i<=m; i++) {
		int a, b; cin>>a>>b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	}
	sort(ord.begin(), ord.end());

	int v;
	for(auto vv : ord) {
		v = vv.se;
		done[v] = 1;

		S.clear();
		for(auto it : adj[v]) if(done[it]) S.insert(find(it));

		summ[v] = B[v];
		for(auto it : S) summ[v] += summ[it];

		DP[v] = summ[v] + vv.fi;
		for(auto it : S) {
			DP[v] = min(DP[v], summ[v]-summ[it] + max(ll(vv.fi), DP[it]));
			Union(v, it);
		}
	}

	cout<<DP[v]<<"\n";
	return 0;
}