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;
#define maxn 100010
#define ll long long
#define db double
#define vi vector<int>
#define pb push_back
#define mod 998244353
#define fi first
#define pi pair<int, int>
#define se second
#define mp make_pair
ll ksm(ll a, ll b) {
   if (!b) return 1;
   ll ns = ksm(a, b >> 1);
   ns = ns * ns % mod;
   if (b & 1) ns = ns * a % mod;
   return ns;
}
// head
int fa[maxn];
ll sz[maxn], ans[maxn];
int gfa(int a) {
	if (fa[a] == a) return a;
	return fa[a] = gfa(fa[a]);
}
int n, m;
vi eg[maxn];
int a[maxn], b[maxn];
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		scanf("%d%d", &a[i], &b[i]), 
		a[i] -= b[i];
	for (int i = 1; i <= m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		eg[u].pb(v), eg[v].pb(u);
	}
	vector<pi> r(n);
	for (int i = 1; i <= n; i++)
		r[i - 1] = mp(a[i], i), 
		sz[i] = b[i];
	sort(r.begin(), r.end());
	for (int i = 0; i < n; i++) {
		int nid = r[i].se;
		fa[nid] = nid;
		ll nans = max(0, a[nid]);
		for (auto v : eg[nid]) {
			if (!fa[v]) continue;
			else {
				int nf = gfa(v);
				if (nf == nid) continue;
				fa[nf] = nid;
				nans = min(nans, max(ans[nf], a[nid] - sz[nf]));
				sz[nid] += sz[nf];
			}
		}
		ans[nid] = nans;
	}
	cout << ans[gfa(1)] + sz[gfa(1)] << endl;
	return 0;
}
