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<iostream>
#include<vector>
#include<algorithm>
using namespace std;

const int N = 100100;
int a[N], b[N], n, m, vis[N], fa[N], c[N];
typedef long long ll;
vector<int>g[N];
ll sum[N], mn[N];
bool cmp(const int& x, const int& y) {
	if (a[x] != a[y])return a[x] < a[y];
	return b[x] > b[y];
}
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) { cin >> a[i] >> b[i]; c[i] = i; }
	sort(c + 1, c + n + 1, cmp);
	for (int i = 1; i <= m; ++i) {
		int x, y; cin >> x >> y;
		g[x].push_back(y); g[y].push_back(x);
	}
	for (int i = 1; i <= n; ++i)fa[i] = i;
	for (int i = 1; i <= n; ++i) {
		int u = c[i]; sum[u] = b[u]; mn[u] = max(0, a[u] - b[u]);
		vis[u] = 1;
		for (int j : g[u])if (vis[j]) {
			int v = find(j); if (u == v)continue;
			sum[u] += sum[v]; mn[u] = min(mn[u], max(a[u] - sum[v] - b[u], mn[v]));
			fa[v] = u;
		}
	}
	cout << sum[c[n]] + mn[c[n]] <<endl;

	return 0;
}