#include<bits/stdc++.h>
using namespace std;
const int MAXN=2e5+10;
int b[MAXN];
int val[MAXN*4];
int dp[MAXN];
int build(int l,int r,int idx){
	if(l==r){
		val[idx]=1e9;
		return val[idx];
	}
	int mid=(l+r)>>1;
	return val[idx]=min(build(l,mid,idx*2),build(mid+1,r,idx*2+1));
}
int modify(int l,int r,int idx,int x,int mod){
	if(l==r)
		return val[idx]=min(val[idx],mod);
	int mid=(l+r)>>1;
	if(x<=mid)
		return val[idx]=min(val[idx],modify(l,mid,idx*2,x,mod));
	else
		return val[idx]=min(val[idx],modify(mid+1,r,idx*2+1,x,mod));
}

int query(int l,int r,int idx,int L,int R){
	if(l>R||r<L)	return 1e9;
	if(L<=l&&r<=R)	return val[idx];
	int mid=(l+r)>>1;
	return min(query(l,mid,idx*2,L,R),query(mid+1,r,idx*2+1,L,R));
}
vector<int> event[MAXN];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	int base=0;
	int cnt=0;
	for(int i=1;i<=n;i++){
		cin>>b[i];
		if(!b[i])	cnt++;
	}
	int q;
	cin>>q;
	for(int i=1,l,r;i<=q;i++){
		cin>>l>>r;
		event[l].push_back(r);
	}
	build(1,n+1,1);
	modify(1,n+1,1,1,0);
	for(int i=1;i<=n;i++){
		sort(event[i].begin(),event[i].end(),greater<int>());
		for(int r:event[i])
			modify(1,n+1,1,r+1,query(1,n+1,1,i,r));
		modify(1,n+1,1,i+1,query(1,n+1,1,i,i)+(b[i]==0?-1:1));
	}
	cout<<cnt+query(1,n+1,1,n+1,n+1)<<endl;
}

### 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;
const int MAXN=2e5+10;
int b[MAXN];
int val[MAXN*4];
int dp[MAXN];
int build(int l,int r,int idx){
	if(l==r){
		val[idx]=1e9;
		return val[idx];
	}
	int mid=(l+r)>>1;
	return val[idx]=min(build(l,mid,idx*2),build(mid+1,r,idx*2+1));
}
int modify(int l,int r,int idx,int x,int mod){
	if(l==r)
		return val[idx]=min(val[idx],mod);
	int mid=(l+r)>>1;
	if(x<=mid)
		return val[idx]=min(val[idx],modify(l,mid,idx*2,x,mod));
	else
		return val[idx]=min(val[idx],modify(mid+1,r,idx*2+1,x,mod));
}

int query(int l,int r,int idx,int L,int R){
	if(l>R||r<L)	return 1e9;
	if(L<=l&&r<=R)	return val[idx];
	int mid=(l+r)>>1;
	return min(query(l,mid,idx*2,L,R),query(mid+1,r,idx*2+1,L,R));
}
vector<int> event[MAXN];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	int base=0;
	int cnt=0;
	for(int i=1;i<=n;i++){
		cin>>b[i];
		if(!b[i])	cnt++;
	}
	int q;
	cin>>q;
	for(int i=1,l,r;i<=q;i++){
		cin>>l>>r;
		event[l].push_back(r);
	}
	build(1,n+1,1);
	modify(1,n+1,1,1,0);
	for(int i=1;i<=n;i++){
		sort(event[i].begin(),event[i].end(),greater<int>());
		for(int r:event[i])
			modify(1,n+1,1,r+1,query(1,n+1,1,i,r));
		modify(1,n+1,1,i+1,query(1,n+1,1,i,i)+(b[i]==0?-1:1));
	}
	cout<<cnt+query(1,n+1,1,n+1,n+1)<<endl;
}

```