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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using Number = System.Int64;
namespace Program {
    public class Solver {
        Random rnd = new Random(0);
        public void Solve() {
            //逆からみる
            //始点含め、初めて訪れると B_v 円もらえる
            //A_v 円払ったら出られる
            //なので入れる条件は C_v = max(0,A_v - B_v) 円もっていること
            //始点を固定してシミュレーションすることにする
            //はじめ X 円、いける頂点が減ることはない
            //O(N^2 polylogN) でできた

            // X 円持ってるときに行き放題な領域、これは森で表現できる
            // C_v の小さいほうから見てC_v>=C_u なら辺を張る、みたいな感じになっている
            // 親のときは C_v 円でやりたい放題
            // 子のときは max(dp[u],C_v-S_u) 円でやりたい放題

            var n = ri; var m = ri;
            var AB = Enumerate(n, x => new long[] { rl, rl });
            var C = Enumerate(n, x => Max(0, AB[x][0] - AB[x][1]));
            var G = Enumerate(n, x => new List<int>());
            foreach (var i in Rep(m)) {
                var u = ri - 1;
                var v = ri - 1;
                G[u].Add(v); G[v].Add(u);
            }
            var H = Enumerate(n, x => new List<int>());
            var r = Enumerate(n, x => x);
            var s = new DisjointSet(n);
            foreach (var id in Enumerate(n, x => x).OrderBy(x => C[x] * 1000000L + x)) {
                foreach (var t in G[id]) {
                    var v = C[id] * 1000000L + id;
                    var u = C[t] * 1000000L + t;
                    if (v >= u && !s.IsUnited(id, t)) {
                        H[id].Add(r[s[t]]);
                        s.Unite(id, t);
                        r[s[id]] = id;
                    }
                }
            }
            Func<int, KeyValuePair<long, long>> dfs = null;
            dfs = (cur) => {
                var sum = AB[cur][1];
                var ret = C[cur];
                foreach (var t in H[cur]) {
                    var res = dfs(t);
                    sum += res.Key;
                    ret = Min(ret, Max(res.Value, C[cur] - res.Key));
                }

                return new KeyValuePair<long, long>(sum, ret);
            };
            var ans = dfs(r[s[0]]);
            Console.WriteLine(ans.Key + ans.Value);
        }
        const long INF = 1L << 60;
        int ri => sc.Integer();
        long rl => sc.Long();
        double rd => sc.Double();
        string rs => sc.Scan();
        public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

        static IEnumerable<int> Rep(int n) => Enumerable.Range(0, n);
        static IEnumerable<int> RRep(int n) => Enumerable.Range(0, n).Reverse();
        static T[] Enumerate<T>(int n, Func<int, T> f) {
            var a = new T[n];
            for (int i = 0; i < a.Length; ++i) a[i] = f(i);
            return a;
        }
        static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
    }
}
#region main
static class Ex {
    static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
    static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
        return string.Join(st, ie);
    }
    static public void Main() {
        Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
        var solver = new Program.Solver();
        solver.Solve();
        Console.Out.Flush();
    }
}
#endregion
#region Ex
namespace Program.IO {
    using System.IO;
    using System.Text;
    using System.Globalization;

    public class Printer : StreamWriter {
        public override IFormatProvider FormatProvider => CultureInfo.InvariantCulture;
        public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    }

    public class StreamScanner {
        public StreamScanner(Stream stream) { str = stream; }

        public readonly Stream str;
        private readonly byte[] buf = new byte[1024];
        private int len, ptr;
        public bool isEof = false;

        private byte read() {
            if (isEof) return 0;
            if (ptr >= len) {
                ptr = 0;
                if ((len = str.Read(buf, 0, 1024)) <= 0) {
                    isEof = true;
                    return 0;
                }
            }
            return buf[ptr++];
        }

        public char Char() {
            byte b = 0;
            do b = read(); while ((b < 33 || 126 < b) && !isEof);
            return (char)b;
        }
        public string Scan() {
            var sb = new StringBuilder();
            for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
            return sb.ToString();
        }
        public string ScanLine() {
            var sb = new StringBuilder();
            for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
            return sb.ToString();
        }
        public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
        public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
        public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
    }
}

#endregion
#region DisjointSet
public class DisjointSet {
    int[] par;
    byte[] rank;
    public DisjointSet(int n) {
        par = new int[n];
        for (int i = 0; i < n; i++)
            par[i] = -1;
        rank = new byte[n];
    }
    public int this[int id] {
        get {
            if ((par[id] < 0)) return id;
            return par[id] = this[par[id]];
        }
    }
    public bool Unite(int x, int y) {
        x = this[x]; y = this[y];
        if (x == y) return false;
        if (rank[x] < rank[y]) { var z = x; x = y; y = z; }
        par[x] += par[y];
        par[y] = x;
        if (rank[x] == rank[y])
            rank[x]++;
        return true;
    }
    public int Size(int x) { return -par[this[x]]; }
    public bool IsUnited(int x, int y) { return this[x] == this[y]; }

}
#endregion