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;

typedef long long ll;
typedef pair<int, int> pii;

#define F first
#define S second

const int MAXN = 1e5 + 10;

int n, m, a[MAXN], b[MAXN], sec[MAXN];
vector<int> adj[MAXN];
int par[MAXN];
ll ans[MAXN], sumB[MAXN];
pair<ll, ll> gec[MAXN];

bool cmp(int u, int v){return a[u]-b[u] < a[v]-b[v];}

int getPar(int v){return par[v]==v? v: par[v]=getPar(par[v]);}

bool mark[MAXN];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < n; i++) cin >> a[i] >> b[i], a[i] = max(a[i], b[i]);
	while (m--){
		int u, v; cin >> u >> v, u--, v--;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	iota(sec, sec + n, 0);
	sort(sec, sec + n, cmp);

	iota(par, par + n, 0);
	for (int i = 0; i < n; i++){
		int v = sec[i];
		mark[v] = true;
		sumB[v] = b[v];

		int tt = 0;
		for (int u:adj[v]){
			if (!mark[u]) continue;
			int pu = getPar(u);
			if (pu == v) continue;

			gec[tt++] = {ans[pu], sumB[pu]};
			par[pu] = v;
			sumB[v] += sumB[pu];
		}

		ans[v] = sumB[v] + (a[v] - b[v]);
		for (int j = 0; j < tt; j++){
			auto x = gec[j];
			ans[v] = min(ans[v], max((ll)a[v], x.F+b[v]) + (sumB[v] - b[v] - x.S));
		}
	}

	cout << ans[sec[n-1]] << "\n";
	return 0;
}
