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;
#define LL long long
const int N = 1e5 + 11;
int n, m;
int fa[N], id[N];
LL a[N], b[N], c[N], sum[N], f[N];
bool vis[N];
vector<int> to[N], son;
int find(int x){
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}
bool cmp(int x, int y){
	return c[x] < c[y];
}
int main(){
	cin>>n>>m;
	for(int i = 1;i <= n; i++){
		scanf("%lld%lld", &a[i], &b[i]);
		c[i] = max(0ll, a[i] - b[i]);
	}
	int u, v;
	for(int i = 1;i <= m; i++){
		scanf("%d%d", &u, &v);
		to[u].push_back(v);
		to[v].push_back(u);
	}
	for(int i = 1;i <= n; i++){
		fa[i] = i, id[i] = i;
	}
	sort(id + 1, id + 1 + n, cmp);
	for(int i = 1;i <= n; i++){
		int u = id[i];
		int sz = to[u].size();
		sum[u] = b[u];
		vis[u] = 1;
		son.clear();
 		for(int j = 0;j < sz; j++){
			int v = to[u][j];
			int fu = find(u), fv = find(v);
			if(fu == fv || !vis[v])continue;
			sum[u] += sum[fv];
			fa[fv] = fu;
			son.push_back(fv);
		}
		f[u] = c[u] + sum[u];
		sz = son.size();
		for(int j = 0;j < sz; j++){
			int v = son[j];
			f[u] = min(f[u], sum[u] - sum[v] + max(c[u], f[v]));
		}
	}
	cout<<f[id[n]]<<endl;
	return 0;
}
