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 X first
#define Y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const ll N = 100005, inf = 2e9;

ll n, m, a[N], b[N], inv[N], s;

vector<ll> adj[N];
pll o[N];

ll p[N], v[N];
bool c[N];

ll Find (ll X) {
	if(p[X] == X) return X;
	return p[X] = Find(p[X]);
}

bool can (ll X) {
	for(ll i=1;i<=n;i++) {
		p[i] = i;
		v[i] = b[i];
	}
	for(ll i=1;i<=n;i++) {
		ll I = o[i].Y;
		c[I] = (a[I]-b[I] <= X);
		for(auto &T : adj[I]) {
			if(inv[T] >= i) continue;
			T = Find(T);
			if(T == I) continue;
			if(c[T] && a[I]-b[I] <= v[T]+X) c[I] = true;
			v[I] += v[T];
			p[T] = I;
		}
	}
	return c[Find(1)];
}

int main()
{
	scanf("%lld%lld",&n,&m);
	for(ll i=1;i<=n;i++) {
		scanf("%lld%lld",&a[i],&b[i]);
		o[i] = {a[i]-b[i], i};
		s += b[i];
	}
	sort(o+1, o+1+n);
	for(ll i=1;i<=n;i++) {
		inv[o[i].Y] = i;
	}
	for(ll i=1;i<=m;i++) {
		ll A, B;
		scanf("%lld%lld",&A,&B);
		adj[A].push_back(B);
		adj[B].push_back(A);
	}
	ll S = 0, E = inf;
	while(S<E) {
		ll M = (S+E)/2;
		can(M) ? E = M : S = M+1;
	}
	printf("%lld\n",S+s);
}
