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>
#define eb emplace_back
using namespace std;
typedef long long ll;

const int MAXN = 100055;

ll dp[MAXN], ds[MAXN];
int ud[MAXN];

vector<int> G[MAXN];
int A[MAXN], B[MAXN], O[MAXN];
bitset<MAXN> chk;

int N, M;

int uf(int i) { return i == ud[i] ? i : (ud[i] = uf(ud[i])); }
void uf(int a, int b) {
	b = uf(b);
	if(uf(a) == b) return;
	ll t = max(dp[b] + ds[b], ll(A[a])) - ds[b];
	a = uf(a);
	ds[a] += ds[b];
	if(t < dp[a]) dp[a] = t;
	ud[b] = a;
}

int main() {
	ios::sync_with_stdio(false);

	cin >> N >> M;
	for(int i = 1; i <= N; i++) {
		cin >> A[i] >> B[i];
		A[i] = max(0, A[i] - B[i]);
	}
	for(int i = 0, a, b; i < M; i++) {
		cin >> a >> b;
		G[a].eb(b);
		G[b].eb(a);
	}

	iota(O, O+N+1, 0); sort(O+1, O+N+1, [&](int a, int b) {
		return A[a] < A[b];
	});

	iota(ud, ud+N+1, 0);
	for(int oi = 1, i; oi <= N; oi++) {
		i = O[oi];
		dp[i] = A[i];
		ds[i] = B[i];
		chk[i] = true;
		for(int v : G[i]) if(chk[v]) uf(i, v);
	}

	cout << dp[uf(1)] + ds[uf(1)] << endl;
	return 0;
}