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;

#define ll  long long
#define ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 2e5 + 5;

typedef pair<ll,ll> ii;

int a[N];
int b[N];
int vis[N];

int p[N];
int c[N];

int lead(int x) {   return  p[x] == 0 ? x : p[x] = lead(p[x]);  }

vector<int> g[N];

ii  dfs(int u)  {
    vector<ii>  vec;

    for(int v : g[u])
        vec.pb(dfs(v));
    
    ll  S = b[u];
    ll  F = g[u].empty() ? max(a[u],b[u]) : 1e18;

    for(ii  T : vec)    S += T.X;
    for(ii  T : vec)    {
        ll  nxt = max(T.Y,1ll * c[u]) + S - T.X;

        if (F > nxt)
            F = nxt;
    }
    return  ii(S,F);
}

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;  cin >> n;
    int m;  cin >> m;

    for(int i = 1 ; i <= n ; ++i)   {
        cin >> a[i];
        cin >> b[i];
        c[i] = a[i] - b[i];

        if (c[i] < 0)
            c[i] = 0;
    }
    vector<int> perm(n);

    iota(all(perm),1);
    sort(all(perm),[&](int x,int y) {   return  c[x] < c[y];    });

    for(int i = 1 ; i <= m ; ++i)   {
        int x;  cin >> x;
        int y;  cin >> y;

        g[x].pb(y);
        g[y].pb(x);
    }
    for(int x : perm)   {
        vector<int> vec = g[x]; g[x].clear();
        vis[x] = 1;

        for(int y : vec)    {
            y = lead(y);
            if (vis[y] && y != x)   {
                g[x].pb(y);
                p[y] = x;
            }
        }
    }
    cout << dfs(perm[n - 1]).Y << endl;
}