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;
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto x : v) cerr << x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int i = 0; i < (n); ++i) cerr << a[i] << ", "; cerr << endl
typedef long long ll;
typedef long double ld;
const int MAXN = 100005;
int A[MAXN], B[MAXN], C[MAXN];
vector<int> E[MAXN];
bool used[MAXN];
int par[MAXN];
ll sumb[MAXN], dp[MAXN];
int root(int a) {
	if (par[a] == a) return a;
	return par[a] = root(par[a]);
}
int main() {
	int N, M;
	scanf("%d%d", &N, &M);
	vector<pair<int, int>> order;
	for (int i = 1; i <= N; ++i) {
		scanf("%d%d", &A[i], &B[i]);
		C[i] = max(0, A[i] - B[i]);
		order.push_back({C[i], i});
		par[i] = i;
	}
	for (int i = 0; i < M; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	sort(order.begin(), order.end());
	for (auto ord : order) {
		int v = ord.second;
		set<int> trees;
		sumb[v] = B[v];
		for (int w : E[v]) {
			if (!used[w]) continue;
			int pw = root(w);
			if (trees.count(pw)) continue;
			sumb[v] += sumb[pw];
			trees.insert(pw);
		}
		used[v] = true;
		dp[v] = C[v] + sumb[v];
		for (int w : trees) {
			ll amt = sumb[v] - sumb[w] + max(dp[w], (ll)C[v]);
			dp[v] = min(dp[v], amt);
			par[w] = v;
		}
	}
	printf("%lld\n", dp[root(1)]);
}
