#include<bits/stdc++.h>
using namespace std;
void getmax(int &x,int y){x=max(x,y);}
void getmin(int &x,int y){x=min(x,y);}
int n,q,l,r;
int b[200100],dat[524288],mx=262144,dp[200100];
int w[200100],pre[200100];
vector<int> opr[200100];
void update(int l,int r,int delta){
	l+=mx;r+=mx;
	while(l<=r){
		getmin(dat[l],delta);
		getmin(dat[r],delta);
		l=(l+1)>>1;
		r=(r-1)>>1;
	}
}
int get(int p){
	p+=mx;
	int ret=1e9;
	while(p){
		getmin(ret,dat[p]);
		p>>=1;
	}
	return ret;
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	for(int i=1;i<mx*2;i++)
		dat[i]=1e9;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>w[i];
		pre[i]=pre[i-1]+(1-w[i]);
	}
	cin>>q;
	for(int i=1;i<=q;i++){
		cin>>l>>r;
		opr[r].push_back(l); 
	}
	for(int i=1;i<=n;i++){
		update(i-1,i-1,dp[i-1]-pre[i-1]);
		dp[i]=dp[i-1]+w[i];
		for(int j:opr[i]){
			update(j,i,get(j-1));
		}
		getmin(dp[i],get(i)+pre[i]);
	}
	cout<<dp[n]<<endl;
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void getmax(int &x,int y){x=max(x,y);}
void getmin(int &x,int y){x=min(x,y);}
int n,q,l,r;
int b[200100],dat[524288],mx=262144,dp[200100];
int w[200100],pre[200100];
vector<int> opr[200100];
void update(int l,int r,int delta){
	l+=mx;r+=mx;
	while(l<=r){
		getmin(dat[l],delta);
		getmin(dat[r],delta);
		l=(l+1)>>1;
		r=(r-1)>>1;
	}
}
int get(int p){
	p+=mx;
	int ret=1e9;
	while(p){
		getmin(ret,dat[p]);
		p>>=1;
	}
	return ret;
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	for(int i=1;i<mx*2;i++)
		dat[i]=1e9;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>w[i];
		pre[i]=pre[i-1]+(1-w[i]);
	}
	cin>>q;
	for(int i=1;i<=q;i++){
		cin>>l>>r;
		opr[r].push_back(l); 
	}
	for(int i=1;i<=n;i++){
		update(i-1,i-1,dp[i-1]-pre[i-1]);
		dp[i]=dp[i-1]+w[i];
		for(int j:opr[i]){
			update(j,i,get(j-1));
		}
		getmin(dp[i],get(i)+pre[i]);
	}
	cout<<dp[n]<<endl;
	return 0;
}

```