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;

const int N = 100005;

int n, m;
int a[N], b[N];
int par[N];
bool vis[N];
long long sum[N], res[N];
vector<int> G[N];

int find(int u) { return u == par[u] ? u : par[u] = find(par[u]); }

void join(int u, int v) { par[find(u)] = find(v); }

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		cin >> a[i] >> b[i];
		par[i] = i, sum[i] = b[i], res[i] = max(a[i], b[i]);
	}
	for (int i = 1; i <= m; ++i) {
		int u, v; cin >> u >> v;
		G[u].push_back(v), G[v].push_back(u);
	}
	vector<int> vec;
	for (int i = 1; i <= n; ++i) vec.push_back(i);
	sort(vec.begin(), vec.end(), [&] (int x, int y) {
		return a[x] - b[x] < a[y] - b[y];
	});
	for (auto i : vec) {
		set<int> s;
		for (auto j : G[i]) {
			if (vis[j]) s.insert(find(j));
		}
		for (auto j : s) {
			join(j, i), sum[i] += sum[j];
		}
		if (s.size()) {
			long long mn = 1e18;
			for (auto j : s) {
				mn = min(mn, max(0LL + a[i] - b[i], res[j]) + sum[i] - sum[j]);
			}
			res[i] = mn;
		}
		vis[i] = 1;
	}
	cout << res[find(vec.back())];
}