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 PII pair<int, int>
#define MP make_pair
#define PB push_back
#define FF first
#define SS second
using namespace std;
template<typename T> inline void gmax(T &x, T y){x = x > y ? x : y;}
template<typename T> inline void gmin(T &x, T y){x = x < y ? x : y;}
const int MAXN = 100010;
int n, m;
LL a[MAXN], b[MAXN], v[MAXN];
int id[MAXN];
vector<int> G[MAXN];
inline bool cmp(const int &a, const int &b){return v[a] < v[b];}
int f[MAXN];
inline int find(int x){return f[x] == x ? f[x] : f[x] = find(f[x]);}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) scanf("%lld%lld", &a[i], &b[i]), v[i] = max(a[i] - b[i], 0LL), id[i] = i;
	sort(id + 1, id + n + 1, cmp);
	for(int i = 1; i <= m; i++){
		int x, y;
		scanf("%d%d", &x, &y);
		G[x].PB(y);
		G[y].PB(x);
	}
	for(int i = 1; i <= n; i++){
		int x = id[i];
		f[x] = x;
		LL tmp = v[x];
		for(unsigned j = 0; j < G[x].size(); j++){
			int y = G[x][j];
			if(!f[y] || (y = find(y)) == x) continue;
			f[y] = x;
			b[x] += b[y];
			gmin(v[x], max(v[y], tmp) - b[y]); 
		}
		v[x] += b[x];
	}
	printf("%lld\n", v[id[n]]);
	return 0;
}