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: Wrong Answer)

using System;
using System.Collections.Generic;

namespace ARC98_F {
    class Program {

        static int[] A;
        static int[] B;
        static int[] AmB;
        static List<Con> lcon = new List<Con>();


        static void Main(string[] args) {
            String line;
            line = Console.ReadLine();
            String[] var = line.Split(new char[] { ' ' });
            var N = int.Parse(var[0]);
            var M = int.Parse(var[1]);

            A = new int[N];
            B = new int[N];
            AmB = new int[N];
            group = new int[N];
            for (int i = 0; i < N; i++) {
                line = Console.ReadLine();
                var = line.Split(new char[] { ' ' });
                A[i] = int.Parse(var[0]);
                B[i] = int.Parse(var[1]);
                AmB[i] = A[i] - B[i];
                if (AmB[i] < 0) { AmB[i] = 0; }
                MakeSet(i);
            }

            for (int i = 0; i < M; i++) {
                line = Console.ReadLine();
                var = line.Split(new char[] { ' ' });
                var U = int.Parse(var[0]) - 1;
                var V = int.Parse(var[1]) - 1;
                if (U > V) { int k = U; U = V; V = k; }
                lcon.Add(new Con(U, V, AmB[U], AmB[V]));
            }

            lcon.Sort((a, b) => (a.C - b.C));

            var tree = new List<int>[N];
            for (int i = 0; i < N; i++) { tree[i] = new List<int>(); }
            var sumB = new int[N];
            for (int i = 0; i < N; i++) { sumB[i] = B[i]; }

            var ss = new List<KeyValuePair<int, int>>();
            for (int i = 0; i < N; i++) { ss.Add(new KeyValuePair<int, int>(-1, i)); }

            for (int i = 0; i < M; i++) {
                var u = lcon[i].U;
                var v = lcon[i].V;
                var gu = Find(u);
                var gv = Find(v);
                if (gu == gv) continue;
                if (AmB[u] < AmB[v]) { var k = u; u = v; v = k; k = gu; gu = gv; gv = k; }
                sumB[gu] += sumB[gv];
                tree[gu].Add(gv);
                Union(u, v);
                ss[gu] = new KeyValuePair<int, int>(i, gu);
            }

            ss.Sort((a, b) => a.Key - b.Key);

            var ans = new int[N];
            for (int i = 0; i < N; i++) { ans[i] = AmB[i] + B[i]; }

            int mq = 0;
            for (int i = 0; i < N; i++) {
                var l = ss[i].Value;
                mq = AmB[l] + sumB[l];
                for (int j = 0; j < tree[l].Count; j++) {
                    var k = tree[l][j];
                    var q = (ans[k] > AmB[l]) ? ans[k] : AmB[l];
                    q = q + sumB[l] - sumB[k];
                    if (q < mq) { mq = q; }
                }
                ans[l] = mq;
            }

            Console.WriteLine(mq.ToString());
        }



        static int[] group;
        static int Find(int x) {
            if (group[x] == x) return x;
            return Find(group[x]);
        }
        static void Union(int x, int y) {
            x = Find(x);
            y = Find(y);
            if (x != y) group[y] = x;
        }
        static void MakeSet(int x) {
            group[x] = x;
        }

    }

    class Con {
        public int U;
        public int V;
        public int C;

        public Con(int u, int v, int c1, int c2) {
            U = u;
            V = v;
            if (c1 > c2) {
                C = c1;
            } else {
                C = c2;
            }
        }
    }

}