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;
const int N = 100005;
int n, m, A[N], B[N], C[N], id[N], dsu[N];
long long dp[N], sum[N];
bool vis[N];
vector<int> G[N];
int Find(int x) { return x == dsu[x] ? x : dsu[x] = Find(dsu[x]); }
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &A[i], &B[i]);
		C[i] = max(A[i] - B[i], 0);
		id[i] = i;
		dsu[i] = i;
		vis[i] = false;
	}
	for (int i = 1; i <= m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	sort(id + 1, id + 1 + n, [](int i, int j) {
		return C[i] < C[j];
	});
	for (int i = 1; i <= n; i++) {
		int u = id[i];
		vis[u] = true;
		sum[u] = B[u];
		vector<int> sons;
		for (int v : G[u]) {
			if (!vis[v] || Find(u) == Find(v)) continue;
			int rv = Find(v);
			sons.push_back(rv);
			sum[u] += sum[rv];
			dsu[rv] = u;
		}
		dp[u] = sum[u] + C[u];
		for (int v : sons) {
			dp[u] = min(dp[u], sum[u] - sum[v] + max(dp[v], 1ll * C[u]));
		}
	}
	printf("%lld\n", dp[id[n]]);
	return 0;
}