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 = 100005;
int n, m, cnt, top, s, maxi, now, fa[Maxn], node[Maxn], sta[Maxn], a[Maxn], b[Maxn], w[Maxn], head[Maxn];
long long sum[Maxn], mini[Maxn];
vector <int> Ve[Maxn];
bool vis[Maxn];
struct edg
{
	int nxt, to;
}edge[2 * Maxn];
void add(int x, int y)
{
	edge[++cnt] = (edg){head[x], y};
	head[x] = cnt;
}
void dfs(int u)
{
	mini[u] = 0x3f3f3f3f3f3f3f3f;
	sum[u] = b[u];
	for (vector <int> :: iterator it = Ve[u].begin(); it != Ve[u].end(); it++)
	{
		int to = *it;
		dfs(to);
		sum[u] += sum[to];
		mini[u] = min(mini[u], max((long long) w[u], mini[to]) - sum[to]);
	}
	if (!Ve[u].size()) mini[u] = w[u] + b[u];
	else mini[u] += sum[u];
}
int get_fa(int x)
{
	return x == fa[x] ? x : fa[x] = get_fa(fa[x]);
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		scanf("%d%d", &a[i], &b[i]), w[i] = max(0, a[i] - b[i]), node[i] = fa[i] = i;
	for (int i = 1; i <= m; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	sort(node + 1, node + 1 + n, [](int x, int y){return w[x] < w[y];});
	for (int j = 1; j <= n; j++)
	{
		int u = node[j];
		vis[u] = true;
		for (int i = head[u]; i; i = edge[i].nxt)
		{
			int to = edge[i].to;
			if (vis[to] && get_fa(to) != u)
				Ve[u].push_back(get_fa(to)), fa[get_fa(to)] = u;
		}
	}
	dfs(node[n]);
	printf("%lld", mini[node[n]]);
	return 0;
}
