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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

#[allow(dead_code)]
struct UnionFind {
    parent: Vec<i32>,
}

#[allow(dead_code)]
impl UnionFind {
    fn new(n: usize) -> UnionFind {
        UnionFind {
            parent: vec![-1; n],
        }
    }
    fn root(&mut self, x: usize) -> usize {
        let p = self.parent[x];
        return if p < 0 {
            x
        } else {
            self.parent[x] = self.root(p as usize) as i32;
            self.parent[x] as usize
        };
    }
    fn union_set(&mut self, x: usize, y: usize) -> bool {
        let mut x = self.root(x);
        let mut y = self.root(y);
        if x == y {
            return false;
        }
        if self.parent[y] < self.parent[x] {
            std::mem::swap(&mut x, &mut y);
        }
        self.parent[x] += self.parent[y];
        self.parent[y] = x as i32;
        true
    }
    fn find_set(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }
    fn size(&mut self, x: usize) -> usize {
        let r = self.root(x);
        -self.parent[r] as usize
    }
}

#[test]
fn union_find() {
    let mut ufind = UnionFind::new(10);
    assert!(ufind.root(0) == 0);
    assert!(ufind.size(3) == 1);
    assert!(ufind.union_set(1, 2));
    assert!(!ufind.union_set(1, 2));
    ufind.union_set(2, 3);
    let r = ufind.root(3);
    assert!(r == 1 || r == 2 || r == 3);
    assert!(ufind.size(3) == 3);
}

fn main() {
    let n = read::<usize>();
    let m = read::<usize>();
    let mut aseq = vec![0i64; n];
    let mut bseq = vec![0i64; n];
    let mut cseq = vec![0i64; n];
    for i in 0..n {
        aseq[i] = read::<i64>();
        bseq[i] = read::<i64>();
        cseq[i] = std::cmp::max(0, aseq[i] - bseq[i]);
    }
    let mut edges = vec![(0, 0, 0); m];
    for i in 0..m {
        let u = read::<usize>() - 1;
        let v = read::<usize>() - 1;
        edges[i] = (std::cmp::max(cseq[u], cseq[v]), u, v);
    }
    edges.sort();
    let mut ufind = UnionFind::new(n);
    let mut dp = vec![0; n];
    let mut bsums = vec![0; n];
    let mut cmaxs = vec![0; n];
    for i in 0..n {
        dp[i] = bseq[i] + cseq[i];
        bsums[i] = bseq[i];
        cmaxs[i] = cseq[i];
    }
    let mut ans = 0;
    for i in 0..m {
        let (_c, mut u, mut v) = edges[i];
        u = ufind.root(u);
        v = ufind.root(v);
        if u == v {
            continue;
        }
        let ndp1 = std::cmp::max(dp[u], cmaxs[v]) + bsums[v];
        let ndp2 = std::cmp::max(dp[v], cmaxs[u]) + bsums[u];
        ufind.union_set(u, v);
        let r = ufind.root(u);
        dp[r] = *[ndp1, ndp2].iter().min().unwrap();
        cmaxs[r] = std::cmp::max(cmaxs[u], cmaxs[v]);
        bsums[r] = bsums[u] + bsums[v];
        ans = std::cmp::max(ans, dp[r]);
    }
    println!("{}", ans);
}