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>

#define LL long long
#define RG register

using namespace std;
template<class T> T gi() {
	T x = 0; bool f = 0; char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = 1, c = getchar();
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f ? -x : x;
}
const int N = 1e5 + 10;
int a[N], b[N], fa[N], p[N];
LL f[N], siz[N];
vector<int> e[N];
bool vis[N];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	int n = gi<int>(), m = gi<int>();
	for (int i = 1; i <= n; i++) a[i] = gi<int>(), siz[i] = b[i] = gi<int>();
	for (int i = 1; i <= m; i++) {
		int x = gi<int>(), y = gi<int>();
		e[x].push_back(y), e[y].push_back(x);
	}
	for (int i = 1; i <= n; i++) fa[i] = p[i] = i, a[i] = max(0, a[i] - b[i]);
	sort(p + 1, p + 1 + n, cmp);
	for (int i = 1; i <= n; i++) {
		int u = p[i]; vector<int> son; vis[u] = 1;		
		for (auto v : e[u]) {
			if (!vis[v] || u == find(v)) continue;
			son.push_back(find(v));
			siz[u] += siz[find(v)]; fa[find(v)] = u;
		}
		f[u] = siz[u] + a[u];
		for (auto v : son) f[u] = min(f[u], siz[u] - siz[v] + max(1ll * a[u], f[v]));
	}
	printf("%lld\n", f[p[n]]);
	return 0;
}
