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
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 100000;
vector<int> g[N + 1];
int n, m, a[N + 1], b[N + 1], pa[N + 1], ex[N + 1];
ll sm[N + 1];
pair<int, int> x[N];

int P(int v) { return pa[v] < 0 ? v : pa[v] = P(pa[v]); }

int main(){
	scanf("%d%d", &n, &m);
	f(i, 1, n + 1)scanf("%d%d", a + i, b + i);
	f(i, 0, n)x[i] = make_pair(a[i + 1] - b[i + 1], i + 1);
	sort(x, x + n);
	f(i, 0, m){
		int a, b;
		scanf("%d%d", &a, &b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	pa[0] = -1;
	f(i, 0, n){
		int v = x[i].second, need = a[v] - b[v];
		sm[v] = b[v];
		ex[v] = max(0, need);
		pa[v] = -1;
		f(j, 0, g[v].size()){
			int u = P(g[v][j]);
			if (u == 0 || u == v)continue;
			sm[v] += sm[u];
			ex[v] = min((ll)ex[v], max((ll)ex[u], need - sm[u]));
			pa[u] = v;
		}
	}
	printf("%lld\n", sm[x[n - 1].second] + ex[x[n - 1].second]);
}