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 For(i,l,r) for(int i = (l),i##end = (r);i <= i##end;i++)
#define Fordown(i,r,l) for(int i = (r),i##end = (l);i >= i##end;i--)
#define debug(x) cout << #x << " = " << x << endl

using namespace std;

typedef long long ll;

template <typename T> inline bool chkmin(T &x,T y) { return y < x ? x = y,1 : 0; }
template <typename T> inline bool chkmax(T &x,T y) { return x < y ? x = y,1 : 0; }

const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;

int A[N],B[N],fa[N],id[N],to[N << 1],nex[N << 1],beg[N],vis[N];
ll Sum[N],dp[N];
int e;

bool Cmp(const int &lhs,const int &rhs) {
	return A[lhs] < A[rhs];
}

inline void add(int x,int y) {
	to[++e] = y,nex[e] = beg[x],beg[x] = e;
}

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

inline int read() {
	int x = 0,flag = 1;
	char ch = getchar();
	while(!isdigit(ch) && ch != '-')ch = getchar();
	if(ch == '-')flag = -1,ch = getchar();
	while(isdigit(ch))x = (x << 3) + (x << 1) + (ch - '0'),ch = getchar();
	return x * flag;
}

int main() {

	int n = read(),m = read();
	For(i,1,n) {
		A[i] = read(),B[i] = read(),A[i] = max(A[i] - B[i],0);
		fa[i] = id[i] = i,Sum[i] = B[i],dp[i] = A[i];
	}
	For(i,1,m) {
		int x = read(),y = read();
		add(x,y),add(y,x);
	}
	sort(id + 1,id + n + 1,Cmp);
	For(i,1,n) {
		vis[id[i]] = true;
		for(int j = beg[id[i]];j;j = nex[j]) {
			if(!vis[to[j]]) continue;
			int x = find(id[i]),y = find(to[j]);
			if(x == y) continue;
			fa[y] = x,Sum[x] += Sum[y];
			chkmin(dp[x],max(dp[y],A[x] - Sum[y]));
		}
	}

	printf("%lld\n",dp[id[n]] + Sum[id[n]]);

	return 0;
}