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;

namespace TYC
{
	typedef long long ll;
	const int N = 1e5 + 5;

	int n, m, A[N], B[N], C[N], Fa[N], P[N], Vis[N];
	ll F[N], Sum[N];
	vector<int> E[N], T[N];

	bool cmp(const int a, const int b) { return C[a] < C[b]; }

	int find(const int a) { return a == Fa[a] ? a : Fa[a] = find(Fa[a]); }

	void dfs(const int u)
	{
		Sum[u] = B[u];
		for (int v : T[u])
		{
			dfs(v);
			Sum[u] += Sum[v];
		}
		if (!T[u].size())
			F[u] = C[u] + B[u];
		else 
		{
			F[u] = LLONG_MAX;
			for (int v : T[u])
				F[u] = min(F[u], max(F[v], (ll)C[u]) + Sum[u] - Sum[v]); 
		}
	}

	void work()
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &A[i], &B[i]);
			C[i] = max(A[i] - B[i], 0);
			P[i] = Fa[i] = i;
		}
		for (int i = 1, u, v; i <= m; i++)
		{
			scanf("%d%d", &u, &v);
			E[u].push_back(v);
			E[v].push_back(u);
		}
		sort(P + 1, P + 1 + n, cmp);
		for (int i = 1; i <= n; i++)
		{
			int u = P[i], f;
			Vis[u] = 1;
			for (int v : E[u])
				if (Vis[v] && (f = find(v))	!= u)
				{
					Fa[f] = u;
					T[u].push_back(f);
				}
		}
		dfs(P[n]);
		printf("%lld\n", F[P[n]]);
	}
}

int main()
{
	TYC::work();
	return 0;
}