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 endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 100005;
int n, m, q[MX], u, v, p[MX];
ll a[MX], b[MX], dp[MX], s[MX];
vi adj[MX], to[MX];
bitset<MX> bs;
int parent (int a) { return p[a] == a ? a : p[a] = parent(p[a]); }

void dfs (int u) {
	for (int v : to[u]) dfs(v), s[u] += s[v];
	s[u] += b[u], dp[u] = to[u].size() ? 1e18 : a[u] + b[u];
	for (int v : to[u]) dp[u] = min(dp[u], s[u] - s[v] + max(dp[v], a[u]));
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	iota(p, p+MX, 0);
	iota(q, q+MX, 1);

	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i] >> b[i];
		a[i] = max(a[i] - b[i], 0ll);
	}

	while (m--) {
		cin >> u >> v;
		adj[u].pb(v);
		adj[v].pb(u);
	}

	sort(q, q+n, [&] (int i, int j) {
		return a[i] < a[j];
	});

	for (int i = 0; i < n; i++) {
		u = q[i];
		bs[u] = 1; 
		for (int v : adj[u]) {
			if (parent(u) == parent(v) || !bs[v]) continue;
			v = parent(v);
			to[u].pb(v);
			p[v] = u;
		}
	}

	dfs(q[n-1]);
	cout << dp[q[n-1]] << endl;

	return 0;
}
