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 maxn = 1e5+10;

int n, m, a[maxn], b[maxn], p[maxn];
vector<int> to[maxn];

bool cmd(int i, int j) { return a[i]-b[i] < a[j]-b[j]; }

int f[maxn];
long long B[maxn], w[maxn];
int find(int x) { return f[x]>0?f[x]=find(f[x]):x; }

bool vis[maxn];

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", a+i, b+i);
		B[i] = b[i];
		p[i] = i;
	}
	int u, v;
	for (int j = 1; j <= m; ++j) {
		scanf("%d%d", &u, &v);
		to[u].push_back(v);
		to[v].push_back(u);
	}
	sort(p+1, p+1+n, cmd);
	for (int _ = 1; _ <= n; ++_) {
		int i = p[_];
//		printf("%d: %d\n", i, a[i]-b[i]);
		w[i] = max(0, a[i]-b[i]);
		for (auto __: to[i]) if (vis[__]) {
			int j = find(__);
			if (j == i) continue;
			if (B[j]+w[j] >= a[i]-b[i]) w[i] = min(w[i], w[j]);
			else w[i] = min(w[i], (a[i]-b[i])-B[j]);
			B[i] += B[j];
			f[j] = i;
		}
		vis[i] = 1;
//		printf("%d: B=%d w=%d\n", i, B[i], w[i]);
	}
	cout << w[p[n]]+B[p[n]] << endl;
}
