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>
#define MAX_N 100000
using namespace std;
typedef long long lnt;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, m; lnt s[MAX_N+5], f[MAX_N+5];
vector <int> G[MAX_N+5]; bool mrk[MAX_N+5];
int a[MAX_N+5], b[MAX_N+5], p[MAX_N+5], fa[MAX_N+5];
bool cmp(const int &x, const int &y) {return a[x] < a[y];}
int getf(int x) {return fa[x] == x ? x : fa[x] = getf(fa[x]);}
int main() {
	read(n), read(m);
	for (int i = 1; i <= n; i++)
		read(a[i]), read(b[i]), a[i] = max(a[i]-b[i], 0);
	for (int i = 1, u, v; i <= m; i++)
		read(u), read(v), G[u].push_back(v), G[v].push_back(u);
	for (int i = 1; i <= n; i++)
		p[i] = fa[i] = i, f[i] = a[i], s[i] = b[i];
	sort(p+1, p+n+1, cmp);
	for (int i = 1; i <= n; i++) {
		int u, fu, fv; mrk[u = p[i]] = true;
		for (int v : G[u]) if (mrk[v] && ((fu = getf(u)) ^ (fv = getf(v))))
			fa[fv] = fu, s[fu] += s[fv], f[fu] = min(f[fu], max(f[fv], a[fu]-s[fv]));
	}
	return printf("%lld\n", f[getf(1)]+s[getf(1)]), 0;
}
