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;
int n, m, x, y;
vector<int> a[100020];
long long A[100020];
long long B[100020];
long long g[100020];
int f[100020];
int r[100020];
int v[100020];
int F(int x) {
	return f[x] != x ? f[x] = F(f[x]) : x;
}
int cmp(int x, int y) {
	return A[x] < A[y];
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%lld%lld", &A[i], &B[i]);
		A[i] = max(A[i] - B[i], 0LL);
		r[i] = i;
	}
	sort(r + 1, r + 1 + n, cmp);
	for (int i = 0; i < m; i++) {
		scanf("%d%d", &x, &y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for (int i = 1; i <= n; i++) {
		v[r[i]] = 1;
		f[r[i]] = r[i];
		g[r[i]] = A[r[i]];
		for (int j: a[r[i]]) {
			j = F(j);
			if (v[j] && j != r[i]) {
				B[r[i]] += B[j];
				g[r[i]] = min(g[r[i]], max(g[j], A[r[i]] - B[j]));
				f[j] = r[i];
			}
		}
	}
	printf("%lld\n", g[F(1)] + B[F(1)]);
	return 0;
}