#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

struct pp
{
	int l,r;
}q[200001];

int f[600001],n,Q,b[200001],sum=0;

bool cmp(pp o,pp p) {return (o.l<p.l)||((o.l==p.l)&&(o.r>p.r));}

int get(int o,int p,int q,int r,int s)
{
	if((o==q)&&(p==r)) return f[s];
	int mid=(o+p)/2;
	if(r<=mid) return get(o,mid,q,r,s*2);else
	if(q>mid) return get(mid+1,p,q,r,s*2+1);else
	return min(get(o,mid,q,mid,s*2),get(mid+1,p,mid+1,r,s*2+1));
}

void put(int o,int p,int q,int r,int s)
{
	f[r]=min(f[r],s);
	if((o==q)&&(p==q)) return;
	int mid=(o+p)/2;
	if(q<=mid) put(o,mid,q,r*2,s);else put(mid+1,p,q,r*2+1,s);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		sum+=b[i]==0;
	}
	memset(f,0,sizeof(f));
	scanf("%d",&Q);
	for(int i=1;i<=Q;i++) scanf("%d%d",&q[i].l,&q[i].r);
	memset(f,13,sizeof(f));
	put(0,n,0,1,0);
	sort(q+1,q+Q+1,cmp);
	int j=1;
	for(int i=1;i<=n;i++)
	{
		while(q[j].l==i) 
		{
			int yu=get(0,n,i-1,q[j].r,1);
			put(0,n,q[j].r,1,yu);
			j++;
		}
		int yt=get(0,n,i-1,i-1,1)+((b[i]==1)?1:(-1));
		put(0,n,i,1,yt);
	}
	cout<<(get(0,n,n,n,1)+sum)<<endl;
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

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


Output

3


Input

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


Output

5


Input

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


Output

1

### Response

```cpp
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

struct pp
{
	int l,r;
}q[200001];

int f[600001],n,Q,b[200001],sum=0;

bool cmp(pp o,pp p) {return (o.l<p.l)||((o.l==p.l)&&(o.r>p.r));}

int get(int o,int p,int q,int r,int s)
{
	if((o==q)&&(p==r)) return f[s];
	int mid=(o+p)/2;
	if(r<=mid) return get(o,mid,q,r,s*2);else
	if(q>mid) return get(mid+1,p,q,r,s*2+1);else
	return min(get(o,mid,q,mid,s*2),get(mid+1,p,mid+1,r,s*2+1));
}

void put(int o,int p,int q,int r,int s)
{
	f[r]=min(f[r],s);
	if((o==q)&&(p==q)) return;
	int mid=(o+p)/2;
	if(q<=mid) put(o,mid,q,r*2,s);else put(mid+1,p,q,r*2+1,s);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		sum+=b[i]==0;
	}
	memset(f,0,sizeof(f));
	scanf("%d",&Q);
	for(int i=1;i<=Q;i++) scanf("%d%d",&q[i].l,&q[i].r);
	memset(f,13,sizeof(f));
	put(0,n,0,1,0);
	sort(q+1,q+Q+1,cmp);
	int j=1;
	for(int i=1;i<=n;i++)
	{
		while(q[j].l==i) 
		{
			int yu=get(0,n,i-1,q[j].r,1);
			put(0,n,q[j].r,1,yu);
			j++;
		}
		int yt=get(0,n,i-1,i-1,1)+((b[i]==1)?1:(-1));
		put(0,n,i,1,yt);
	}
	cout<<(get(0,n,n,n,1)+sum)<<endl;
	return 0;
}
```