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 <cstdio>
#include <algorithm>
#include <vector>

typedef long long LL;
const LL Infll = 0x3f3f3f3f3f3f3f3f;
const int MN = 100005;

int N, M, A[MN], B[MN], P[MN], vis[MN];
std::vector<int> G[MN], T[MN];

int fa[MN];
inline int ff(int x) { return fa[x] ? fa[x] = ff(fa[x]) : x; }

LL f[MN], s[MN];

void DFS(int u) {
	s[u] = B[u], f[u] = Infll;
	for (auto v : T[u]) DFS(v), s[u] += s[v];
	for (auto v : T[u]) f[u] = std::min(f[u], s[u] - s[v] + std::max((LL)A[u], f[v]));
	if (T[u].empty()) f[u] = A[u] + B[u];
}

int main() {
	scanf("%d%d", &N, &M);
	for (int i = 1; i <= N; ++i)
		scanf("%d%d", &A[i], &B[i]),
		A[i] = std::max(A[i] - B[i], 0),
		P[i] = i;
	for (int i = 1, x, y; i <= M; ++i)
		scanf("%d%d", &x, &y),
		G[x].push_back(y),
		G[y].push_back(x);
	std::sort(P + 1, P + N + 1, [](int i, int j) { return A[i] < A[j]; });
	for (int i = 1; i <= N; ++i) {
		int u = P[i];
		for (auto v : G[u])
			if (vis[v] && ff(v) != u)
				T[u].push_back(ff(v)),
				fa[ff(v)] = u;
		vis[u] = 1;
	}
	DFS(P[N]);
	printf("%lld\n", f[P[N]]);
	return 0;
}