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 = 100100;

int n,m,a[maxn],b[maxn],p[maxn],id[maxn],vis[maxn],root;
ll f[maxn],s[maxn];
vector<int> g[maxn];

int cmp(int i,int j) {
    return a[i] < a[j];
}

int get(int x) {
    return ((x==p[x]) ? x : p[x] = get(p[x]));
}

int main() {
   // freopen("in.txt","r",stdin);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    int ma =0;
    for (int i=1;i<=n;i++) {
        cin>>a[i]>>b[i];
        a[i] = max(a[i]-b[i],0); // all of node always need atleast ai-bi
        if (a[i] > ma) {
            ma = a[i];
            root = i;
        }
        id[i] = i;
        p[i] = i;
        s[i] = b[i]; // amount of money need to donate
        f[i] = a[i]; // amount of money need to add more to donate all
    }
    for (int i=1;i<=m;i++) {
        int x,y;
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    sort(id+1,id+n+1,cmp);
    for (int i=1;i<=n;i++) {
        int u = id[i];
        vis[u] = 1;
        for (auto v : g[u]) if (vis[v]) { // alway keep order to DP
            int x = get(u), y= get(v);
           // cout<<u<<" "<<v<<" "<<x<<" "<<y<<endl;
            if (x==y) continue;
            p[y] = x; s[x] += s[y];
            ll need = a[u] - s[y];
            f[x] = min(f[x],max(need,f[y]));
        }
    }
    cout<<f[root] + s[root]<<endl;
}
