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;
const int maxn = 100000 + 10;

int n, m, a[maxn], b[maxn], h[maxn], ecnt, pos[maxn], sfa[maxn], vis[maxn], w[maxn];
ll f[maxn], dp[maxn];
struct enode{
	int v, n;
	enode() {}
	enode(int _v, int _n):v(_v), n(_n) {}
}e[maxn << 1];

inline void addedge(int u, int v) { ecnt ++; e[ecnt] = enode(v,h[u]); h[u] = ecnt; }

inline bool cmp(int x, int y) { return w[x] < w[y]; }

int findfa(int u) { return (sfa[u] == u) ? (u) : (sfa[u] = findfa(sfa[u])); }

int lis[maxn];

inline void upmin(ll &x, ll y) { x = min(x,y); }

int main() {
	scanf("%d%d", &n, &m); ecnt = 0;
	memset(h,-1,sizeof(h));
	for(int i = 1;i <= n;i ++) scanf("%d%d", &a[i], &b[i]), w[i] = max(0,a[i] - b[i]);
	for(int i = 1;i <= m;i ++) {
		int u, v;
		scanf("%d%d", &u, &v);
		addedge(u,v);
		addedge(v,u);
	}
	for(int i = 1;i <= n;i ++) pos[i] = i, sfa[i] = i;
	sort(pos + 1,pos + n + 1,cmp);
	for(int i = 1;i <= n;i ++) {
		int u = pos[i]; vis[u] = 1;
		f[u] = b[u]; int tot = 0;
		for(int j = h[u];~ j;j = e[j].n) {
			int v = e[j].v, r;
			if(!vis[v] || findfa(v) == findfa(u)) continue;
			r = findfa(v);
			f[u] += f[r];
			sfa[r] = u;
			lis[++ tot] = r;
		}
		dp[u] = f[u] + w[u];
		for(int j = 1;j <= tot;j ++) {
			int v = lis[j];
			ll now = f[u] - f[v] + max((ll)w[u],dp[v]);
			upmin(dp[u],now);
		}
	}
	ll Ans = dp[pos[n]];
	printf("%lld\n", Ans);
	return 0;
}
