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, z[100005], par[100005], tb, tc;
long long x[100005], y[100005], sum[100005], dp[100005], ta;
vector< int > v[100005], vv[100005];

int find(int a) {
	return par[a] == a ? a : par[a] = find(par[a]);
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%lld%lld", &ta, x + i);
		y[i] = max(ta - x[i], 0LL);
		z[i] = i;
	}
	sort(z + 1, z + n + 1, [](const int &a, const int &b) {return y[a] < y[b];});
	while (m--) {
		scanf("%d%d", &tb, &tc);
		v[tb].push_back(tc);
		v[tc].push_back(tb);
	}
	for (int i = 1; i <= n; i++) {
		tb = z[i];
		par[tb] = tb;
		sum[tb] = x[tb];
		dp[tb] = y[tb];
		for (int j: v[tb])
			if (par[j] && (j = find(j)) != tb) {
				vv[tb].push_back(j);
				par[j] = tb;
				sum[tb] += sum[j];
				dp[tb] = min(dp[tb], max(dp[j], y[tb]) - sum[j]);
			}
		dp[tb] += sum[tb];
	}
	printf("%lld\n", dp[z[n]]);
	return 0;
}
