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 pb push_back
#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 1e5 + 10;

typedef long long LL;

vector<int> G[N];
int fa[N], rk[N], n, m;
int A[N], B[N];
LL dp[N], sum[N];

int find(int x) {
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

bool cmp(int x, int y) {
	return A[x] - B[x] < A[y] - B[y]; 
}

int main() {

	scanf("%d%d", &n, &m);
	For(i, 1, n) rk[i] = i, scanf("%d%d", &A[i], &B[i]);
	For(i, 1, m) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v), G[v].pb(u);
	}
	
	sort(rk + 1, rk + n + 1, cmp);
	For(i, 1, n) {
		int o = rk[i];
		fa[o] = o;
		dp[o] = max(0, A[o] - B[o]), sum[o] = B[o];
		for (int v : G[o]) {
			if (!fa[v] || find(v) == o) continue;
			int x = find(v);
			fa[x] = o;
			dp[o] = min(dp[o], max(dp[x], A[o] - B[o] - sum[x]));
			sum[o] += sum[x];
		}
	}
	printf("%lld\n", dp[find(1)] + sum[find(1)]);

	return 0;
}