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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// need
#include <iostream>
#include <algorithm>

// data structure
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <complex>
//#include <deque>
#include <valarray>
#include <unordered_map>

// stream
//#include <istream>
//#include <sstream>
//#include <ostream>
#include <fstream>

// etc
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>

// input
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
#define VAR(type, ...)type __VA_ARGS__;MACRO_VAR_Scan(__VA_ARGS__);
template<typename T> void MACRO_VAR_Scan(T& t) { std::cin >> t; }
template<typename First, typename...Rest>void MACRO_VAR_Scan(First& first, Rest&...rest) { std::cin >> first; MACRO_VAR_Scan(rest...); }
#define VEC_ROW(type, n, ...)std::vector<type> __VA_ARGS__;MACRO_VEC_ROW_Init(n, __VA_ARGS__); for(int i=0; i<n; ++i){MACRO_VEC_ROW_Scan(i, __VA_ARGS__);}
template<typename T> void MACRO_VEC_ROW_Init(int n, T& t) { t.resize(n); }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Init(int n, First& first, Rest&...rest) { first.resize(n); MACRO_VEC_ROW_Init(n, rest...); }
template<typename T> void MACRO_VEC_ROW_Scan(int p, T& t) { std::cin >> t[p]; }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Scan(int p, First& first, Rest&...rest) { std::cin >> first[p]; MACRO_VEC_ROW_Scan(p, rest...); }
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define MAT(type, c, m, n) std::vector<std::vector<type>> c(m, std::vector<type>(n));for(auto& r:c)for(auto& i:r)std::cin>>i;

// output
#define OUT(d) std::cout<<(d);
#define FOUT(n, d) std::cout<<std::fixed<<std::setprecision(n)<<(d);
#define SOUT(n, c, d) std::cout<<std::setw(n)<<std::setfill(c)<<(d);
#define SP std::cout<<" ";
#define TAB std::cout<<"\t";
#define BR std::cout<<"\n";
#define SPBR(i, n) std::cout<<(i + 1 == n ? '\n' : ' ');
#define ENDL std::cout<<std::endl;
#define FLUSH std::cout<<std::flush;
#define SHOW(d) {std::cerr << #d << "\t:" << (d) << "\n";}
#define SHOWVECTOR(v) {std::cerr << #v << "\t:";for(const auto& xxx : v){std::cerr << xxx << " ";}std::cerr << "\n";}
#define SHOWVECTOR2(v) {std::cerr << #v << "\t:\n";for(const auto& xxx : v){for(const auto& yyy : xxx){std::cerr << yyy << " ";}std::cerr << "\n";}}
#define SHOWQUEUE(a) {auto tmp(a);std::cerr << #a << "\t:";while(!tmp.empty()){std::cerr << tmp.front() << " ";tmp.pop();}std::cerr << "\n";}

// utility
#define ALL(a) (a).begin(),(a).end()
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define RFOR(i, a, b) for(int i=(b)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define FORLL(i, a, b) for(ll i=ll(a);i<ll(b);++i)
#define RFORLL(i, a, b) for(ll i=ll(b)-1;i>=ll(a);--i)
#define REPLL(i, n) for(ll i=0;i<ll(n);++i)
#define RREPLL(i, n) for(ll i=ll(n)-1;i>=0;--i)
#define IN(a, x, b) (a<=x && x<b)
template<typename T> inline T CHMAX(T& a, const T b) { return a = (a < b) ? b : a; }
template<typename T> inline T CHMIN(T& a, const T b) { return a = (a > b) ? b : a; }
#define EXCEPTION(msg) throw std::string("Exception : " msg " [ in ") + __func__ + " : " + std::to_string(__LINE__) + " lines ]"
#define TRY(cond, msg) try {if (cond) EXCEPTION(msg);}catch (std::string s) {std::cerr << s << std::endl;}
void CHECKTIME(std::function<void()> f) { auto start = std::chrono::system_clock::now(); f(); auto end = std::chrono::system_clock::now(); auto res = std::chrono::duration_cast<std::chrono::nanoseconds>((end - start)).count(); std::cerr << "[Time:" << res << "ns  (" << res / (1.0e9) << "s)]\n"; }

// test
template<class T> std::vector<std::vector<T>> VV(int n, int m, T init = T()) {
	return std::vector<std::vector<T>>(n, std::vector<T>(m, init));
}
template<typename S, typename T>
std::ostream& operator<<(std::ostream& os, std::pair<S, T> p) {
	os << "(" << p.first << ", " << p.second << ")"; return os;
}

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using PAIR = std::pair<int, int>;
using PAIRLL = std::pair<ll, ll>;
constexpr int INFINT = 1 << 30;                          // 1.07x10^ 9
constexpr int INFINT_LIM = (1LL << 31) - 1;              // 2.15x10^ 9
constexpr ll INFLL = 1LL << 60;                          // 1.15x10^18
constexpr ll INFLL_LIM = (1LL << 62) - 1 + (1LL << 62);  // 9.22x10^18
constexpr double EPS = 1e-9;
constexpr int MOD = 1000000007;
constexpr double PI = 3.141592653589793238462643383279;

template<class T, size_t N> void FILL(T(&a)[N], const T& val) { for (auto& x : a) x = val; }
template<class ARY, size_t N, size_t M, class T> void FILL(ARY(&a)[N][M], const T& val) { for (auto& b : a) FILL(b, val); }
template<class T> void FILL(std::vector<T>& a, const T& val) { for (auto& x : a) x = val; }
template<class ARY, class T> void FILL(std::vector<std::vector<ARY>>& a, const T& val) { for (auto& b : a) FILL(b, val); }

// ------------>8------------------------------------->8------------

//Union-Find
class UnionFind {
private:
	std::vector<int> parent;
	std::vector<int> height;
	std::vector<int> m_size;

public:
	UnionFind(int size_) : parent(size_), height(size_, 0), m_size(size_, 1) {
		for (int i = 0; i < size_; ++i) parent[i] = i;
	}
	void init(int size_) {
		parent.resize(size_);
		height.resize(size_, 0);
		m_size.resize(size_, 0);
		for (int i = 0; i < size_; ++i) parent[i] = i;
	}
	int find(int x) {
		if (parent[x] == x) return x;
		return parent[x] = find(parent[x]);
	}
	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y) return;
		int t = size(x) + size(y);
		m_size[x] = m_size[y] = t;
		if (height[x] < height[y]) parent[x] = y;
		else parent[y] = x;
		if (height[x] == height[y]) ++height[x];
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
	int size(int x) {
		if (parent[x] == x) return m_size[x];
		return size(parent[x] = find(parent[x]));
	}
};

signed main() {
	INIT;

	VAR(int, n, m);
	VEC_ROW(int, n, a, b);
	VEC_ROW(int, m, u, v);
	std::vector<std::vector<int>> g(n);
	REP(i, m) {
		--u[i]; --v[i];
		g[u[i]].emplace_back(v[i]);
		g[v[i]].emplace_back(u[i]);
	}
	int bsum = 0;
	REP(i, n) bsum += b[i];

	auto check = [&](int w) -> bool {
		std::priority_queue<PAIR, std::vector<PAIR>, std::greater<PAIR>> pq;
		REP(i, n) pq.push(PAIR(a[i] - b[i], i));
		std::vector<bool> used(n, false);

		std::vector<int> sum(b);
		UnionFind uf(n);

		while (!pq.empty()) {
			PAIR startP = pq.top(); pq.pop();
			int start = startP.second;
			if (used[start]) continue;
			int nowsum = w;

			std::set<PAIR> set;
			set.insert(PAIR(a[start] - b[start], start));
			while (!set.empty()) {
				PAIR nowP = *set.begin(); set.erase(set.begin());
				if (nowsum < nowP.first) break;
				int now = nowP.second;
				used[now] = true;
				if (start != now) {
					int t = sum[uf.find(start)] + sum[uf.find(now)];
					uf.unite(start, now);
					sum[uf.find(start)] = t;
					nowsum = w + t;
				}
				else {
					nowsum = w + b[start];
				}
				for (auto& to : g[now]) if(!uf.same(now, to)) {
					PAIR p = PAIR(a[to] - b[to], to);
					if (!set.count(p)) set.insert(p);
				}
			}
		}
		return uf.size(0) == n;
	};

	int l = -1, r = INFLL;
	while (r - l > 1) {
		int m = (l + r) / 2;
		if (check(m)) r = m;
		else l = m;
	}
	OUT(r + bsum)BR;
	return 0;
}
/*
逆を考える
- 今いる頂点vからB[v]円を回収する．これは各頂点1度しかできない．
- 今いる頂点vに対して所持金がA[v]以上のとき，vに隣接する頂点に移動する．

*/