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 <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxN = 1e5 + 5;
ll a[maxN], b[maxN], f[maxN];
bool vis[maxN];
int N, M, root[maxN], l[maxN], e, idx[maxN];
struct edge{
	int v, las;
}G[maxN << 1];
void add(int u, int v) {
	G[e].v = v; G[e].las = l[u]; l[u] = e++;
}
int getrt(int v) {
	if(root[v] != 0) {
		int rt = getrt(root[v]);
		root[v] = rt; v = rt;
	}
	return v;
}
bool cmp(ll x, ll y) {return a[x] < a[y];}
int main() {
	scanf("%d%d", &N, &M);
	for(int i = 1; i <= N; ++i) 
	scanf("%lld%lld", &a[i], &b[i]), a[i] = max(0ll, a[i] - b[i]), l[i] = -1, idx[i] = i;
	for(int i = 1; i <= M; ++i) {
		int u, v; scanf("%d%d", &u, &v); add(u, v); add(v, u);
	}
	sort(idx + 1, idx + 1 + N, cmp);
	for(int i = 1; i <= N; ++i) {
		int k = idx[i]; f[k] = a[k]; vis[k] = 1;
		for(int j = l[k]; j != -1; j = G[j].las) {
			if(vis[G[j].v]) {
				int v = G[j].v;
				int rt = getrt(v);
				if(rt == k) continue;
				root[rt] = k;
				b[k] += b[rt];
				f[k] = min(f[k], max(f[rt], a[k] - b[rt]));
			}
		}
	}
	cout << b[idx[N]] + f[idx[N]] << endl;
	return 0;
}