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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

#define N	100000
#define M	100000
#define INF	0x3f3f3f3f

int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }

int ds[N], rr[N];

int find(int i) {
	return ds[i] < 0 ? i : (ds[i] = find(ds[i]));
}

int join(int i, int j) {
	i = find(i);
	j = find(j);
	if (i == j)
		return 0;
	if (ds[i] > ds[j]) {
		ds[i] = j;
		rr[j] = rr[i];
	} else {
		if (ds[i] == ds[j])
			ds[i]--;
		ds[j] = i;
	}
	return 1;
}

int aa[N], bb[N];
int ii[M], jj[M];

int compare(const void *a, const void *b) {
	int h1 = *(int *) a;
	int h2 = *(int *) b;
	int i1 = ii[h1];
	int i2 = ii[h2];

	return aa[i1] != aa[i2] ? aa[i1] - aa[i2] : i1 - i2;
}

int main() {
	static int hh[M], ss[N], dp[N];
	int n, m, h, i, j;
	long long ans;

	scanf("%d%d", &n, &m);
	for (i = 0; i < n; i++)
		scanf("%d%d", &aa[i], &bb[i]);
	for (h = 0; h < m; h++) {
		int tmp;

		scanf("%d%d", &i, &j), i--, j--;
		if (aa[i] < aa[j] || aa[i] == aa[j] && i < j)
			tmp = i, i = j, j = tmp;
		ii[h] = i, jj[h] = j, hh[h] = h;
	}
	qsort(hh, m, sizeof *hh, compare);
	for (i = 0; i < n; i++) {
		ds[i] = -1, rr[i] = i;
		ss[i] = bb[i];
	}
	for (h = 0; h < m; h++)
		dp[ii[h]] = INF;
	for (h = 0; h < m; h++) {
		i = ii[hh[h]], j = rr[find(jj[hh[h]])];
		if (join(i, j)) {
			dp[i] = min(dp[i], max(dp[j], aa[i] - (bb[i] + ss[j])));
			ss[i] = min(ss[i] + ss[j], INF);
		}
	}
	ans = dp[rr[find(0)]];
	for (i = 0; i < n; i++)
		ans += bb[i];
	printf("%lld\n", ans);
	return 0;
}