#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
struct tree
{
	int minv[600000],n;
	tree(int size):n(size){memset(minv,0x3f,sizeof minv);}
	void update(int p,int x,int o,int l,int r)
	{
		if(l==r){minv[o]=min(minv[o],x);return;}
		int m=(l+r)>>1;
		if(p<=m)update(p,x,o*2,l,m);
		else update(p,x,o*2+1,m+1,r);
		minv[o]=min(minv[o*2],minv[o*2+1]);
	}
	int query(int x,int y,int o,int l,int r)
	{
		if(x<=l&&r<=y)return minv[o];
		int m=(l+r)>>1,ans=1<<28;
		if(x<=m)ans=min(ans,query(x,y,o*2,l,m));
		if(y>m)ans=min(ans,query(x,y,o*2+1,m+1,r));
		return ans;
	}
	void set(int p,int x){update(p,x,1,0,n);}
	int get(int l,int r){return query(l,r,1,0,n);}
};
vector<int> v[300000];
int f[300000],b[300000],n,q;
int main()
{
	scanf("%d",&n);
	int cnt=0;
	for(int i=1;i<=n;i++)scanf("%d",b+i),cnt+=!b[i],b[i]=b[i]?1:-1;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		v[l].push_back(r);
	}
	tree t(n);
	t.set(0,0);
	for(int i=1;i<=n;i++)
	{
		for(int j:v[i])t.set(j,t.get(i-1,j));
		t.set(i,min(t.get(i,i),t.get(i-1,i-1)+b[i]));
	}
	printf("%d\n",cnt+t.get(n,n));
	return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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<cstdio>
#include<cstring>
#include<vector>
using namespace std;
struct tree
{
	int minv[600000],n;
	tree(int size):n(size){memset(minv,0x3f,sizeof minv);}
	void update(int p,int x,int o,int l,int r)
	{
		if(l==r){minv[o]=min(minv[o],x);return;}
		int m=(l+r)>>1;
		if(p<=m)update(p,x,o*2,l,m);
		else update(p,x,o*2+1,m+1,r);
		minv[o]=min(minv[o*2],minv[o*2+1]);
	}
	int query(int x,int y,int o,int l,int r)
	{
		if(x<=l&&r<=y)return minv[o];
		int m=(l+r)>>1,ans=1<<28;
		if(x<=m)ans=min(ans,query(x,y,o*2,l,m));
		if(y>m)ans=min(ans,query(x,y,o*2+1,m+1,r));
		return ans;
	}
	void set(int p,int x){update(p,x,1,0,n);}
	int get(int l,int r){return query(l,r,1,0,n);}
};
vector<int> v[300000];
int f[300000],b[300000],n,q;
int main()
{
	scanf("%d",&n);
	int cnt=0;
	for(int i=1;i<=n;i++)scanf("%d",b+i),cnt+=!b[i],b[i]=b[i]?1:-1;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		v[l].push_back(r);
	}
	tree t(n);
	t.set(0,0);
	for(int i=1;i<=n;i++)
	{
		for(int j:v[i])t.set(j,t.get(i-1,j));
		t.set(i,min(t.get(i,i),t.get(i-1,i-1)+b[i]));
	}
	printf("%d\n",cnt+t.get(n,n));
	return 0;
}

```