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.

Examples

Input

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4


Output

6


Input

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


Output

44


Input

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


Output

582
#include <bits/stdc++.h>
using namespace std;

inline int gi()
{
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	int sum = 0;
	while('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

const int maxn = 100005;

int n, m, a[maxn], b[maxn], p[maxn], fa[maxn], vis[maxn];
long long f[maxn], s[maxn];

struct edge
{
	int to, next;
} e[maxn * 2];
int h[maxn], tot;

inline void add(int u, int v)
{
	e[++tot] = (edge) {v, h[u]}; h[u] = tot;
	e[++tot] = (edge) {u, h[v]}; h[v] = tot;
}

int find(int x)
{
	if (fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

int main()
{
	n = gi(); m = gi();
	for (int i = 1; i <= n; ++i) {
		a[i] = gi(); b[i] = gi();
		a[i] = max(0, a[i] - b[i]);
	}
	for (int i = 1; i <= m; ++i) add(gi(), gi());
	
	for (int i = 1; i <= n; ++i) fa[i] = p[i] = i, s[i] = b[i], f[i] = a[i];

	sort(p + 1, p + n + 1, [](const int &x, const int &y) {return a[x] < a[y];});

	for (int k = 1; k <= n; ++k) {
		int u = p[k];
		vis[u] = 1;
		for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
			if (vis[v]) {
				int x = find(u), y = find(v);
				if (x == y) continue;
				fa[y] = x;
				s[x] += s[y];
				f[x] = min(f[x], max(f[y], a[x] - s[y]));
			}
	}

	int x = find(1);
	printf("%lld\n", f[x] + s[x]);
	
	return 0;
}

