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 N = 1e5 + 1;
int par[N],n,b[N],m;
long long sum[N],v[N];
vector<int>g[N];
vector<pair<int,int>>temp;
bool vis[N];
int get(int u){return par[u] == u ? u : par[u] = get(par[u]);}
int main(){
	//freopen("readin.txt","r",stdin);
	scanf("%d%d",&n,&m);
	for(int i = 0,a;i < n;i++){
		par[i] = i;
		scanf("%d%d",&a,b + i);
		temp.push_back({a - b[i],i});
	}
	for(int i = 0,a,b;i < m;i++){
		scanf("%d%d",&a,&b);
		g[--a].push_back(--b);
		g[b].push_back(a);
	}
	sort(temp.begin(),temp.end());
	for(auto i : temp){
		int need = max(0,i.first),at = i.second;
		sum[at] = b[at];
		v[at] = need;
		vis[at] = 1;
		for(auto j : g[at]){
			int from = get(j);
			if(from == at || !vis[from])continue;
			sum[at]+=sum[from];
			if(sum[from] + v[from] >= need)v[at] = min(v[at],v[from]);
			else v[at] = min(v[at],need - sum[from]);
			// v[at] = min(v[at],max(v[from],need - sum[from]));
			par[from] = at;
		}
	}
	printf("%lld\n",sum[temp.back().second] + v[temp.back().second]);
}