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 PyPy3 (2.4.0) (judge verdict: Accepted)

import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7


# MOD = 998244353


class UnionFind:
    def __init__(self, size=None, nodes=None):
        """
        size か nodes どっちか指定。
        nodes は set、size は list を使う。
        :param int size:
        :param collections.Iterable nodes:
        """
        assert size is not None or nodes is not None
        if size is not None:
            self._parents = [i for i in range(size)]
            self._ranks = [0 for _ in range(size)]
            self._sizes = [1 for _ in range(size)]
        else:
            self._parents = {k: k for k in nodes}
            self._ranks = {k: 0 for k in nodes}
            self._sizes = {k: 1 for k in nodes}

    def unite(self, x, y):
        """
        x が属する木と y が属する木を併合
        :param x:
        :param y:
        """
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return

        # rank が小さい方が下
        if self._ranks[x] > self._ranks[y]:
            # x が root
            self._parents[y] = x
            self._sizes[x] += self._sizes[y]
        else:
            # y が root
            self._parents[x] = y
            self._sizes[y] += self._sizes[x]
            if self._ranks[x] == self._ranks[y]:
                self._ranks[y] += 1

    def root(self, x):
        """
        x が属する木の root
        :param x:
        """
        if self._parents[x] == x:
            return x
        self._parents[x] = self.root(self._parents[x])
        return self._parents[x]

    def size(self, x):
        """
        x が属する木のノード数
        :param x:
        """
        return self._sizes[self.root(x)]


def argsort(li, key=None, reverse=False):
    return [i for _, i in sorted(
        [(a, i) for i, a in enumerate(li)], key=(lambda t: key(t[0])) if key else None, reverse=reverse
    )]


N, M = list(map(int, sys.stdin.buffer.readline().split()))
A, B = list(zip(*[map(int, sys.stdin.buffer.readline().split()) for _ in range(N)]))
VU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]

# from libs.debug import plot_graph
#
# plot_graph(VU)

graph = [[] for _ in range(N)]
for v, u in VU:
    v -= 1
    u -= 1
    graph[v].append(u)
    graph[u].append(v)

reqs = [max(a, b) for a, b in zip(A, B)]
uses = list(B)

idx = argsort([a - b for a, b in zip(A, B)])
uf = UnionFind(size=N)
seen = [False] * N
ans = [INF] * N
for v in idx:
    seen[v] = True
    roots = set()
    for u in graph[v]:
        if not seen[u]:
            continue
        roots.add(uf.root(u))

    use_all = uses[v]
    for r in roots:
        use_all += uses[r]

    # v が最後
    req_v = reqs[v] + use_all - uses[v]
    for r in roots:
        req_v = min(req_v, max(reqs[r], reqs[v] - uses[v]) + use_all - uses[r])

    for r in roots:
        uf.unite(v, r)
    reqs[uf.root(v)] = req_v
    uses[uf.root(v)] = use_all
ans = reqs[uf.root(0)]
print(ans)