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
// IOI 2021
#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define ends ' '
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << ends << H; debug_out(T...); }
#define debug(...) cerr << "{" << #__VA_ARGS__ << "}:", debug_out(__VA_ARGS__)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll MOD = 1e9 + 7;
 
////////////////////////////////////////////////////////////////////
 
const int N = 1e5 + 5;

int A[N], B[N], P[N];
ll S[N], W[N];
vector< pair<int, pii> > E;

int GetPar(int v) { return P[v] == v ? v : P[v] = GetPar(P[v]); }
void Merge(int v, int u) {
	if (A[v] - B[v] < A[u] - B[u]) swap(v, u);
	int pv = GetPar(v), pu = GetPar(u);
	if (pv == pu) return;
	W[pv] = min(W[pv], max(0LL, (W[pu] + S[pu] >= A[v] - B[v] ? W[pu] : A[v] - B[v] - S[pu])));
	S[pv] += S[pu];
	P[pu] = pv;
}

int main() {
 
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	mt19937 Rnd(time(0));

	
	int n, m; cin >> n >> m;
	for (int i = 1; i <= n; i++) cin >> A[i] >> B[i], P[i] = i, S[i] = B[i], W[i] = max(0, A[i] - B[i]);
	for (int i = 1; i <= m; i++) {
		int v, u; cin >> v >> u;
		E.push_back(make_pair(max(A[v] - B[v], A[u] - B[u]), make_pair(v, u)));
	}
	sort(all(E));
	for (auto e : E) {
		int v = e.second.first, u = e.second.second;
		Merge(v, u);
	}
	cout << max(0LL, W[GetPar(1)] + S[GetPar(1)]) << endl;

	return 0;
}
