#include<bits/stdc++.h>
using namespace std;

const int N=200100,inf=1e9;
int a[N],ans,n,q1,q2,q3,q,mn[N*4];
vector<int>vec[N];

void bt(int o,int l,int r){
	if(l==r){
		mn[o]=(l?inf:0);
		return;
	}
	int mid=(l+r)/2;
	bt(o*2,l,mid);bt(o*2+1,mid+1,r);
	mn[o]=min(mn[o*2],mn[o*2+1]);
}

void query(int o,int l,int r){
	if(q1<=l&&r<=q2){
		q3=min(q3,mn[o]);
		return;
	}
	int mid=(l+r)/2;
	if(q1<=mid)query(o*2,l,mid);
	if(q2>mid)query(o*2+1,mid+1,r);
}

void modify(int o,int l,int r){
	if(l==r){
		mn[o]=min(mn[o],q3);
		return;
	}
	int mid=(l+r)/2;
	if(q1<=mid)modify(o*2,l,mid);
	else modify(o*2+1,mid+1,r);
	mn[o]=min(mn[o*2],mn[o*2+1]);
}


int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>a[i],ans+=!a[i];
	for(cin>>q;q --> 0;){
		int l,r;cin>>l>>r;
		vec[l].push_back(r);
	}
	bt(1,0,n);
	for(int i=1;i<=n;++i){
		for(int j:vec[i]){
			q1=i-1;q2=j;q3=inf;
			query(1,0,n);
			q1=q2=j;
			modify(1,0,n);
		}
		q1=q2=i-1;q3=inf;
		query(1,0,n);
		q1=q2=i;q3+=2*a[i]-1;
		modify(1,0,n);
	}
	q1=q2=n;q3=inf;query(1,0,n);
	cout<<(ans+q3)<<'\n';
	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<bits/stdc++.h>
using namespace std;

const int N=200100,inf=1e9;
int a[N],ans,n,q1,q2,q3,q,mn[N*4];
vector<int>vec[N];

void bt(int o,int l,int r){
	if(l==r){
		mn[o]=(l?inf:0);
		return;
	}
	int mid=(l+r)/2;
	bt(o*2,l,mid);bt(o*2+1,mid+1,r);
	mn[o]=min(mn[o*2],mn[o*2+1]);
}

void query(int o,int l,int r){
	if(q1<=l&&r<=q2){
		q3=min(q3,mn[o]);
		return;
	}
	int mid=(l+r)/2;
	if(q1<=mid)query(o*2,l,mid);
	if(q2>mid)query(o*2+1,mid+1,r);
}

void modify(int o,int l,int r){
	if(l==r){
		mn[o]=min(mn[o],q3);
		return;
	}
	int mid=(l+r)/2;
	if(q1<=mid)modify(o*2,l,mid);
	else modify(o*2+1,mid+1,r);
	mn[o]=min(mn[o*2],mn[o*2+1]);
}


int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>a[i],ans+=!a[i];
	for(cin>>q;q --> 0;){
		int l,r;cin>>l>>r;
		vec[l].push_back(r);
	}
	bt(1,0,n);
	for(int i=1;i<=n;++i){
		for(int j:vec[i]){
			q1=i-1;q2=j;q3=inf;
			query(1,0,n);
			q1=q2=j;
			modify(1,0,n);
		}
		q1=q2=i-1;q3=inf;
		query(1,0,n);
		q1=q2=i;q3+=2*a[i]-1;
		modify(1,0,n);
	}
	q1=q2=n;q3=inf;query(1,0,n);
	cout<<(ans+q3)<<'\n';
	return 0;
}

```