Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4

Sample Output 1

6

If you have 6 yen initially, you can win the game as follows:

Stand on Vertex 4. This is possible since you have not less than 6 yen.

Donate 2 yen to Vertex 4. Now you have 4 yen.

Move to Vertex 3. This is possible since you have not less than 4 yen.

Donate 1 yen to Vertex 3. Now you have 3 yen.

Move to Vertex 2. This is possible since you have not less than 1 yen.

Move to Vertex 1. This is possible since you have not less than 3 yen.

Donate 1 yen to Vertex 1. Now you have 2 yen.

Move to Vertex 2. This is possible since you have not less than 1 yen.

Donate 2 yen to Vertex 2. Now you have 0 yen.

If you have less than 6 yen initially, you cannot win the game. Thus, the answer is 6.

Sample Input 2

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

Sample Output 2

44

Sample Input 3

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

Sample Output 3

582

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5;
int n, m;
ll a[maxn + 10], b[maxn + 10];
vector<int> g[maxn + 10];
int fa[maxn + 10];
ll sum[maxn + 10], f[maxn + 10];
int id[maxn + 10];
bool vis[maxn + 10];

bool cmp(int x, int y) {
	return a[x] < a[y];
}

int getf(int p) {
	return fa[p] == p ? p : fa[p] = getf(fa[p]);
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", &a[i], &b[i]);
		a[i] = max(a[i] - b[i], 0ll);
		sum[i] = b[i]; id[i] = fa[i] = i;
	}
	sort(id + 1, id + n + 1, cmp);
	for (int i = 1; i <= m; ++i) {
		int l, r; scanf("%d%d", &l, &r);
		g[l].push_back(r); g[r].push_back(l);
	}
	for (int i = 1; i <= n; ++i) {
		int p = id[i]; vis[p] = 1;
		bool fd = 1;
		f[p] = 1e18;
		for (int j = 0; j < (int)g[p].size(); ++j) {
			int e = g[p][j];
			if (vis[e]) {
				e = getf(e);
				if (e != p) {
					fa[e] = p; sum[p] += sum[e];
					f[p] = min(f[p], max(a[p], f[e]) - sum[e]);
					fd = 0;
				}
			}
		}
		if (fd) f[p] = 0;
		f[p] += sum[p];
	}
	printf("%lld", f[id[n]]);
}