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;

typedef long long ll;

const int N = 100005;
int n, m, fa[N], c[N];
vector<int> g[N];
ll sumb[N], dp[N];

int find(int x) {
	return !fa[x] ? x : fa[x] = find(fa[x]);
}

int main() {
	scanf("%d%d", &n, &m);
	static int id[N], rk[N];
	for (int i = 1, a, b; i <= n; ++i) {
		scanf("%d%d", &a, &b);
		id[i] = i;
		sumb[i] = b;
		c[i] = max(a - b, 0);
	}
	for (int i = 1, u, v; i <= m; ++i) {
		scanf("%d%d", &u, &v);
		g[u].push_back(v), g[v].push_back(u);
	}
	sort(id + 1, id + 1 + n, [](int x, int y) { return c[x] < c[y]; });
	for (int i = 1; i <= n; ++i)
		rk[id[i]] = i;
	for (int i = 1; i <= n; ++i) {
		dp[id[i]] = c[id[i]];
		for (int v : g[id[i]]) {
			int j = find(v);
			if (rk[v] < i && j != id[i]) {
				fa[j] = id[i];
				sumb[id[i]] += sumb[j];
				dp[id[i]] = min(dp[id[i]], max(dp[j], c[id[i]] - sumb[j]));
			}
		}
	}
	printf("%lld\n", dp[id[n]] + sumb[id[n]]);
	return 0;
}
