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>

typedef long long ll;
const int N = 100054, M = N * 2;

int V, E, Es = 0;
int a[N], d[N], o[N];
int to[M], first[N], next[M];
int p[N];
bool used[N];
ll D[N], f[N];

inline int doz(int x) {return ~(x >> 31) & x;}
inline void down(ll &x, const ll y) {x > y ? x = y : 0;}
inline ll max(const ll x, const ll y) {return x < y ? y : x;}
int ancestor(int x) {return p[x] == x ? x : (p[x] = ancestor(p[x]));}

inline void addedge(int u, int v) {
	to[++Es] = v, next[Es] = first[u], first[u] = Es;
	to[++Es] = u, next[Es] = first[v], first[v] = Es;
}

int main() {
	int i, j, u, v; ll c;
	scanf("%d%d", &V, &E);
	for (i = 1; i <= V; ++i) scanf("%d%d", a + i, d + i), a[i] = doz(a[i] - d[i]), D[i] = d[i];
	for (i = 0; i < E; ++i) scanf("%d%d", &u, &v), addedge(u, v);
	std::iota(p, p + (V + 1), 0), std::iota(o, o + V, 1);
	std::sort(o, o + V, [] (const int x, const int y) {return a[x] < a[y];});
	for (i = 0; i < V; ++i) {
		used[u = o[i]] = true, c = a[u];
		for (j = first[u]; j; j = next[j])
			if (used[to[j]] && (v = ancestor(to[j])) != u)
				p[v] = u, D[u] += D[v], down(c, max(f[v], a[u]) - D[v]);
		f[u] = D[u] + c;
	}
	printf("%lld\n", f[u]);
	return 0;
}