#include <bits/stdc++.h>
#include <iostream>
using namespace std;
using ll = long long;//63bit型整数型
//     モッドを使った累乗
int main(){
//B以上は基本詳細を書いておくと便利　A ＝ ooの値とか　見直す時に便利
// この問題の本質
ll N,M;
cin >> N >> M;
vector<vector<ll>>graph(N,vector<ll>(0));
for(ll i = 0;i<M;i++){
	ll u,v;
	cin >> u >> v;
	u--;
	v--;
	graph[u].push_back(v);
	graph[v].push_back(u);
}
ll L;
cin >> L;
vector<ll>pass(N,-1);
for(ll i = 0;i<L;i++){
	ll J,K;
	cin >> J >> K;
	J--;
	pass[J] = max(pass[J],K);
}
priority_queue<pair<ll,ll>>pq;
for(ll i = 0;i<N;i++)if(pass[i] != -1)pq.push({pass[i],i});
while(true){
	if(pq.empty())break;
	ll distance = pq.top().first;
	ll point = pq.top().second;
	pq.pop();
	for(ll i = 0;i<graph[point].size();i++){
		ll x = graph[point][i];
		if(pass[x] < distance - 1){
			pass[x] = distance - 1;
			pq.push({pass[x],x});
		}
	}
}
ll inf = 1000000000;
vector<ll>ans(N,inf);
vector<bool>visited(N);
queue<pair<ll,ll>>BFS;
if(pass[0] == -1)BFS.push({0,0});
while(true){
	if(BFS.empty())break;
	ll point = BFS.front().first;
	ll count = BFS.front().second;
	ans[point] = min(ans[point],count);
	BFS.pop();
	visited[point] = true;
	for(ll i = 0;i<graph[point].size();i++){
		ll x = graph[point][i];
		if(visited[x] != true && pass[x] == -1){
			BFS.push({x,count+1});
			visited[x] = true;
		}
	}
}
if(visited[N-1] == true){
	cout << "Yes" << "\n";
	cout << ans[N-1] << "\n";
}
else cout << "No" << "\n";
}