#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int N=2e5+5;
int n,b[N],m,sum;
vector<int>vec[N];
int d[N<<2],tag[N<<2];
void build(int l,int r,int o){
	if(l==r)d[o]=0x3fffffff;else{
		const int mid=l+r>>1;
		build(l,mid,o<<1),build(mid+1,r,o<<1|1);
		d[o]=0x3fffffff;
	}
}
void pushdown(int o){
	int&x=tag[o];
	tag[o<<1]+=x,tag[o<<1|1]+=x;
	d[o<<1]+=x,d[o<<1|1]+=x,x=0;
}
void modify(int l,int r,int o,const int&pos,const int&val){
	if(l==r)d[o]=min(d[o],val);else{
		if(tag[o])pushdown(o);
		const int mid=l+r>>1;
		if(pos<=mid)modify(l,mid,o<<1,pos,val);else modify(mid+1,r,o<<1|1,pos,val);
		d[o]=min(d[o<<1],d[o<<1|1]);
	}
}
void modify(int l,int r,int o,const int&L,const int&R,const int&val){
	if(L<=l&&r<=R){
		d[o]+=val,tag[o]+=val;
	}else{
		pushdown(o);
		const int mid=l+r>>1;
		if(L<=mid)modify(l,mid,o<<1,L,R,val);
		if(mid<R)modify(mid+1,r,o<<1|1,L,R,val);
		d[o]=min(d[o<<1],d[o<<1|1]);
	}
}
int query(int l,int r,int o,const int&L,const int&R){
	if(L<=l&&r<=R)return d[o];
	pushdown(o);
	const int mid=l+r>>1;
	if(L<=mid&&mid<R)return min(query(l,mid,o<<1,L,R),query(mid+1,r,o<<1|1,L,R));
	if(L<=mid)return query(l,mid,o<<1,L,R);
	return query(mid+1,r,o<<1|1,L,R);
}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>b[i],sum+=!b[i];
	for(cin>>m;m--;){
		int l,r;
		cin>>l>>r;
		vec[l].push_back(r);
	}
	build(0,n,1);
	modify(0,n,1,0,0);
	for(int i=1;i<=n;++i){
		for(int y:vec[i])modify(0,n,1,y,query(0,n,1,0,y));
		modify(0,n,1,0,i-1,b[i]?1:-1);
	}
	printf("%d\n",query(0,n,1,0,n)+sum);
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int N=2e5+5;
int n,b[N],m,sum;
vector<int>vec[N];
int d[N<<2],tag[N<<2];
void build(int l,int r,int o){
	if(l==r)d[o]=0x3fffffff;else{
		const int mid=l+r>>1;
		build(l,mid,o<<1),build(mid+1,r,o<<1|1);
		d[o]=0x3fffffff;
	}
}
void pushdown(int o){
	int&x=tag[o];
	tag[o<<1]+=x,tag[o<<1|1]+=x;
	d[o<<1]+=x,d[o<<1|1]+=x,x=0;
}
void modify(int l,int r,int o,const int&pos,const int&val){
	if(l==r)d[o]=min(d[o],val);else{
		if(tag[o])pushdown(o);
		const int mid=l+r>>1;
		if(pos<=mid)modify(l,mid,o<<1,pos,val);else modify(mid+1,r,o<<1|1,pos,val);
		d[o]=min(d[o<<1],d[o<<1|1]);
	}
}
void modify(int l,int r,int o,const int&L,const int&R,const int&val){
	if(L<=l&&r<=R){
		d[o]+=val,tag[o]+=val;
	}else{
		pushdown(o);
		const int mid=l+r>>1;
		if(L<=mid)modify(l,mid,o<<1,L,R,val);
		if(mid<R)modify(mid+1,r,o<<1|1,L,R,val);
		d[o]=min(d[o<<1],d[o<<1|1]);
	}
}
int query(int l,int r,int o,const int&L,const int&R){
	if(L<=l&&r<=R)return d[o];
	pushdown(o);
	const int mid=l+r>>1;
	if(L<=mid&&mid<R)return min(query(l,mid,o<<1,L,R),query(mid+1,r,o<<1|1,L,R));
	if(L<=mid)return query(l,mid,o<<1,L,R);
	return query(mid+1,r,o<<1|1,L,R);
}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>b[i],sum+=!b[i];
	for(cin>>m;m--;){
		int l,r;
		cin>>l>>r;
		vec[l].push_back(r);
	}
	build(0,n,1);
	modify(0,n,1,0,0);
	for(int i=1;i<=n;++i){
		for(int y:vec[i])modify(0,n,1,y,query(0,n,1,0,y));
		modify(0,n,1,0,i-1,b[i]?1:-1);
	}
	printf("%d\n",query(0,n,1,0,n)+sum);
	return 0;
}
```