Problem:
Score : 1000 points

Problem Statement

You are given a sequence a = \{a_1, ..., a_N\} with all zeros, and a sequence b = \{b_1, ..., b_N\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.

Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

1 \leq N \leq 200,000

b consists of 0 and 1.

1 \leq Q \leq 200,000

1 \leq l_i \leq r_i \leq N

If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:

N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q

Output

Print the minimum possible hamming distance.

Sample Input 1

3
1 0 1
1
1 3

Sample Output 1

1

If you choose to perform the operation, a will become \{1, 1, 1\}, for a hamming distance of 1.

Sample Input 2

3
1 0 1
2
1 1
3 3

Sample Output 2

0

If both operations are performed, a will become \{1, 0, 1\}, for a hamming distance of 0.

Sample Input 3

3
1 0 1
2
1 1
2 3

Sample Output 3

1

Sample Input 4

5
0 1 0 1 0
1
1 5

Sample Output 4

2

It may be optimal to perform no operation.

Sample Input 5

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7

Sample Output 5

3

Sample Input 6

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13

Sample Output 6

5

Sample Input 7

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9

Sample Output 7

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma comment(linker, "/STACK:60777216")  

#include <cassert>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<complex>
#include<vector>
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<string>
#include<cstdlib>
#include<memory.h>
#include<ctime>
#include<bitset>
#include<fstream>
#include<queue>
#include<stack>
#include<unordered_map>
#include<unordered_set>

using namespace std;

typedef long double ld;

typedef long long ll;
typedef pair<int,int>	pii;
typedef pair<ld,ld>	pdd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef pair<ll,ll> pl;

#define FOR(i,a,b)		for(int i=(a);i<(b);i++)
#define REP(i,n)		FOR(i,0,n)
#define SORT(v)			sort((v).begin(),(v).end())
#define UN(v)			SORT(v),(v).erase(unique((v).begin(),(v).end()),(v).end())
#define CL(a,b)			memset(a,b,sizeof a)
#define pb				push_back	

int n;
int a[222222];
int q;
pii s[222222];

int r[222222];

int main(){	
#ifdef LocalHost
	freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
#endif

	cin>>n;
	REP(i,n)scanf("%d",a+i);
	cin>>q;
	REP(i,q)scanf("%d %d",&s[i].second,&s[i].first),s[i].first--,s[i].second--;
	sort(s,s+q);

	int pos = 0;

	REP(i,n){
		r[i]=(a[i]==0?0:1)+(i>0?r[i-1]:0);
		while(pos<q && s[pos].first==i){
			int from = s[pos].second;
			
			int val = (from>=1?r[from-1]:0);

			for(int j=from;j<=i;++j){
				val += (a[j]?0:1);
				r[j]=min(r[j],val);
			}

			++pos;
		}
	}
	cout<<r[n-1]<<endl;

	/*
	10001111110001
    11111
	 111111111111
	         11111
			 */
#ifdef LocalHost
	printf("TIME: %.3lf\n",ld(clock())/CLOCKS_PER_SEC);
#endif
	return 0;
}