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 (GCC 5.4.1) (judge verdict: Accepted)

#if 0
cat <<EOF >mistaken-paste
#endif
// thx Ebi-chan!

#pragma GCC diagnostic ignored "-Wincompatible-pointer-types"
#define _USE_MATH_DEFINES

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <time.h>

#define BIG 2000000007
#define VERYBIG 20000000000000007LL

#define MOD 1000000007
#define FOD  998244353
typedef uint64_t ull;
typedef  int64_t sll;

#define N_MAX 1048576

#ifdef __cplusplus
#include <queue>
#include <stack>
#include <tuple>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <functional>
#include <array>

using std::queue;
using std::priority_queue;
using std::stack;
using std::tuple;
using std::set;
using std::map;
using std::vector;
using std::greater;
using std::pair;
using std::string;
using std::get;

template <typename T, typename U>
pair<T, U> operator+(pair<T, U> l, pair<T, U> r) {
	return pair<T, U>(l.first + r.first, l.second + r.second);
}

#endif

#ifndef M_PI
#define M_PI 3.14159265358979323846264338327950
#endif

typedef struct {
	sll a;
	sll b;
} hwll;

typedef struct {
	sll a;
	sll b;
	sll c;
} hwllc;

typedef struct {
	hwll a;
	hwll b;
} linell;

typedef struct {
	double a;
	double b;
} hwreal;

sll n, m;
sll h, w;
sll k;
sll q;
sll va, vb, vc, vd, ve, vf;
ull ua, ub, uc, ud, ue, uf;
long double vra, vrb, vrc;
double vda, vdb, vdc;
char ch, dh;

ull umin (ull x, ull y) {
	return (x < y) ? x : y;
}

ull umax (ull x, ull y) {
	return (x > y) ? x : y;
}

sll smin (sll x, sll y) {
	return (x < y) ? x : y;
}

sll smax (sll x, sll y) {
	return (x > y) ? x : y;
}

ull gcd (ull x, ull y) {
	if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}

ull bitpow (ull a, ull x, ull modulo) {
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= modulo;
		}
		x /= 2;
		a = (a * a) % modulo;
	}
	return result;
}

ull divide (ull a, ull b, ull modulo) {
	return (a * bitpow(b, modulo - 2, modulo)) % modulo;
}

ull udiff (ull a, ull b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

sll sdiff (sll a, sll b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

int bitcount (ull n) {
	int result = 0;
	while (n) {
		if (n & 1) result++;
		n /= 2;
	}
	return result;
}

#define BEGCMP(NAME) int32_t NAME (const void *left, const void *right)
#define DEFLR(TYPE) TYPE l=*(TYPE*)left,r=*(TYPE*)right
#define CMPRET(L, R) if((L)<(R))return-1;if((L)>(R))return+1

// int32_t pullcomp (const void *left, const void *right) {
// 	ull l = *(ull*)left;
// 	ull r = *(ull*)right;
// 	if (l < r) {
// 		return -1;
// 	}
// 	if (l > r) {
// 		return +1;
// 	}
// 	return 0;
// }
BEGCMP(pullcomp){
	DEFLR(ull);
	CMPRET(l, r);
	return 0;
}
BEGCMP(prevcomp){
	DEFLR(sll);
	CMPRET(r, l);
	return 0;
}
BEGCMP(psllcomp){
	DEFLR(sll);
	CMPRET(l, r);
	return 0;
}
BEGCMP(pcharcomp){
	DEFLR(char);
	CMPRET(l, r);
	return 0;
}
BEGCMP(pdoublecomp){
	DEFLR(double);
	CMPRET(l, r);
	return 0;
}

int32_t pstrcomp (const void *left, const void *right) {
	char* l = *(char**)left;
	char* r = *(char**)right;

	return strcmp(l, r);
}

BEGCMP(phwllABcomp){
	DEFLR(hwll);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	return 0;
}
BEGCMP(phwllREVcomp){
	DEFLR(hwll);
	CMPRET(l.b, r.b);
	CMPRET(l.a, r.a);
	return 0;
}
BEGCMP(ptriplecomp){
	DEFLR(hwllc);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	CMPRET(l.c, r.c);
	return 0;
}
BEGCMP(ptripleREVcomp){
	DEFLR(hwllc);
	CMPRET(l.b, r.b);
	CMPRET(l.a, r.a);
	CMPRET(l.c, r.c);
	return 0;
}
BEGCMP(ptripleCABcomp){
	DEFLR(hwllc);
	CMPRET(l.c, r.c);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	return 0;
}
BEGCMP(phwrealcomp){
	DEFLR(hwreal);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	return 0;
}

int32_t pquadcomp (const void *left, const void *right) {
	linell l = *(linell*)left;
	linell r = *(linell*)right;

	sll ac = phwllABcomp(&(l.a), &(r.a));
	if (ac) return ac;
	sll bc = phwllABcomp(&(l.b), &(r.b));
	if (bc) return bc;

	return 0;
}
int32_t pfracomp (const void *left, const void *right) {
	hwllc l = *(hwllc*)left;
	hwllc r = *(hwllc*)right;

	CMPRET(l.a * r.b, l.b * r.a);
	return 0;
}
bool isinrange (sll left, sll x, sll right) {
	return (left <= x && x <= right);
}

bool isinrange_soft (sll left, sll x, sll right) {
	return (left <= x && x <= right) || (left >= x && x >= right);
}

void sw (sll *l, sll *r) {
	if (*l == *r) return;
	sll t = *l;
	*l = *r;
	*r = t;
}

ull frac[N_MAX * 3], invf[N_MAX * 3];
ull ncr (sll n, sll r, ull m) {
	if (n < 0 || r < 0 || n < r) return 0;
	return frac[n] * (invf[r] * invf[n - r] % m) % m;
}

sll a[N_MAX + 5];
// ull a[N_MAX + 5];
// sll a[3001][3001];
sll b[N_MAX + 5];
// ull b[N_MAX + 5];
// sll b[3001][3001];
sll c[N_MAX + 5];
sll d[N_MAX + 5];
sll e[N_MAX * 4];
char s[N_MAX + 1];
// char s[3010][3010];
char t[N_MAX + 1];
// char t[3010][3010];
char u[N_MAX + 1];
hwll xy[N_MAX + 5];
hwllc tup[N_MAX + 5];
sll table[3005][3005];
ull gin[N_MAX];
// here we go

hwll sorted[N_MAX];
hwll tg[N_MAX];
ull tin[N_MAX];

ull rank[N_MAX];

ull dp[N_MAX];
ull bsum[N_MAX];
void dfs (ull v) {
	dp[v] = smax(a[v] - b[v], 0);

	for (sll i = tin[v]; i < tin[v + 1]; i++) {
		ull u = tg[i].b;

		dfs(u);
		if (dp[u] + bsum[u] + b[v] >= a[v]) {
			dp[v] = smin(dp[v], dp[u]);
		} else {
			dp[v] = smin(dp[v], a[v] - bsum[u] - b[v]);
		}

		bsum[v] += bsum[u];
	}
	bsum[v] += b[v];

	// printf("%lld: %lld + %lld\n", v, dp[v], bsum[v]);
}

ull parent[N_MAX], size[N_MAX];
void init (ull n) {
	for (sll i = 0; i < n; i++) {
		parent[i] = i;
		size[i] = 1;
	}
}
ull find (ull x) {
	if (parent[x] == x) return x;
	return parent[x] = find(parent[x]);
}
bool unite (ull p, ull v) {
	parent[v] = p;
	return true;
}

ull solve () {
	sll i, j, ki, li;
	// ull result = 0;
	sll result = 0;
	double dresult = 0;
	// ull maybe = 0;
	sll maybe = 0;
	// ull sum = 0;
	sll sum = 0;
	sll item;
	sll *dpcell;

	for (i = 0; i < m; i++) {
		xy[i + m] = (hwll){xy[i].b, xy[i].a};
	}
	qsort(xy, m * 2, sizeof(hwll), phwllABcomp);
	i = j = 0;
	while (i <= n) {
		gin[i] = j;
		while (j < m * 2 && xy[j].a == i) j++;
		i++;
	}

	for (i = 0; i < n; i++) {
		sorted[i] = (hwll){smax(a[i] - b[i], 0), i};
	}
	qsort(sorted, n, sizeof(hwll), phwllABcomp);
	for (i = 0; i < n; i++) {
		rank[sorted[i].b] = i;
		// printf("%lld..\n", sorted[i].b);
	}

	init(n);

	ki = 0;
	for (i = 0; i < n; i++) {
		ull v = sorted[i].b;

		for (j = gin[v]; j < gin[v + 1]; j++) {
			ull u = xy[j].b;

			if (rank[u] >= rank[v]) continue;
			if (find(u) == find(v)) continue;

			ull up = find(u);
			// printf("%lld->%lld\n", v, up);
			tg[ki++] = (hwll){v, up};
			unite(v, up);
		}
	}
	if (ki != n - 1) abort();

	qsort(tg, n - 1, sizeof(hwll), phwllABcomp);
	i = j = 0;
	while (i <= n) {
		tin[i] = j;
		while (j < n - 1 && tg[j].a == i) j++;
		i++;
	}

	ull root = sorted[n - 1].b;
	dfs(root);

	result = dp[root] + bsum[root];

	printf("%lld\n", result);
	// printf("%.15f\n", dresult);
	// puts(s);

	return 0;

	success:
	// puts("YES");
	puts("Yes");
	// printf("%llu\n", result);
	// puts("0");
	// puts("First");
	return 0;

	fail:
	// puts("NO");
	// puts("No");
	// puts("0");
	puts("-1");
	// puts("-1 -1 -1");
	// puts("Second");
	return 1;
}

int32_t main (int argc, char *argv[]) {
	int32_t i, j;

	n = 4;
	m = 0;

	// scanf("%llu", &m);
	// scanf("%llu%llu", &h, &w);
	scanf("%llu%llu", &n, &m);
	// scanf("%llu", &k, &n, &m);
	// scanf("%llu%llu", &h, &w);
	// scanf("%llu", &q);
	// scanf("%lld%lld", &va, &vb, &vc, &vd);
	// va--;
	// vb--;
	// scanf("%llu%llu%llu%llu", &ua, &ub, &uc, &ud, &ue);
	// scanf("%s", s);
	// scanf("%s", t);
	// scanf("%s", u);
	// scanf("%llu", &k);
	// scanf("%lld", &m);
	// for (i = 0; i < n; i++) {
	// 	// scanf("%lld", &a[i]);
	// 	scanf("%lld", &d[i]);
	// }
	// scanf("%llu", &q);
	for (i = 0; i < n; i++) {
		// scanf("%lld%lld", &xy[i].a, &xy[i].b);
		// scanf("%lld%lld%lld", &tup[i].a, &tup[i].b, &tup[i].c);
		// scanf("%lld", &c[i]);

		scanf("%lld", &a[i]);
		scanf("%lld", &b[i]);
		// scanf("%lld", &c[i]);
		// scanf("%lld", &d[i]);
		// a[i]--;
		// b[i]--;
		// c[i]--;
		// d[i]--;
		// xy[i].a--;
		// xy[i].b--;
		// tup[i].a--;
		// tup[i].b--;
	}
	// scanf("%llu", &m);
	// scanf("%llu", &q);
	for (i = 0; i < m; i++) {
		scanf("%lld%lld", &xy[i].a, &xy[i].b);
		xy[i].a--;
		xy[i].b--;
		// scanf("%lld%lld", &a[i], &b[i]);
		// scanf("%lld", &b[i]);
		// a[i]--;
		// b[i]--;
		// scanf("%lld", &c[i]);
		// scanf("%lld", &d[i]);
		// scanf("%lld", &e[i]);
		// c[i]--;
		// d[i]--;
	}

	// for (i = 0; i < q; i++) {
	// 	// scanf("%lld%lld", &xy[i].a, &xy[i].b);
	// 	scanf("%lld", &c[i]);
	// 	// xy[i].a--;
	// 	// xy[i].b--;
	// }

	// for (i = 0; i < h; i++) {
	// 	for (j = 0; j < w; j++) {
	// 		scanf("%lld", &table[i][j]);
	// 		// table[i][j]--;
	// 	}
	// }
	// for (i = 0; i < h; i++) {
	// 	scanf("%s", s[i]);
	// }
	// scanf("%llu", &q);

	solve();

	return 0;
}