#include<bits/stdc++.h>
#define X first
#define Y second
#define N 200005
using namespace std;
vector<int>vec[N];
int T[N<<2],b[N],sum[N],New[N];
int ql,qr,n,i,k,ans,m,Zero;
void build(int x,int l,int r){
	T[x]=-1e9;if (l==r) return;
  int mid=(l+r)>>1;
  build(x<<1,l,mid);
  build(x<<1|1,mid+1,r);
}
int Query(int x,int l,int r){
	if (ql<=l&&r<=qr) return T[x];
	int mid=(l+r)>>1,ret=-1e9;
	if (ql<=mid) ret=max(ret,Query(x<<1,l,mid));
	if (qr>mid)  ret=max(ret,Query(x<<1|1,mid+1,r));
	return ret;
}
void Update(int x,int l,int r,int pos,int val){
	if (l==r) {T[x]=max(T[x],val);return;}
	int mid=(l+r)>>1;
	if (pos<=mid) Update(x<<1,l,mid,pos,val);
	         else Update(x<<1|1,mid+1,r,pos,val);
	T[x]=max(T[x<<1],T[x<<1|1]);
}
int main(){
	scanf("%d",&n);
	for (i=1;i<=n;i++) 
	  scanf("%d",&b[i]),sum[i]=sum[i-1]+b[i];
	scanf("%d",&m);
	for (i=1;i<=m;i++)
	  scanf("%d%d",&ql,&qr),vec[ql].push_back(qr);
	build(1,0,n); 
	Update(1,0,n,0,0-sum[0]);
	for (i=1;i<=n;i++){
		ql=qr=i-1;Zero=Query(1,0,n)+sum[i-1]+(b[i]==0);
		for (k=0;k<vec[i].size();k++){
			ql=i-1;qr=vec[i][k];
			New[k]=sum[vec[i][k]]+Query(1,0,n);
		}
		Update(1,0,n,i,Zero-sum[i]);
		for (k=0;k<vec[i].size();k++)
		  Update(1,0,n,vec[i][k],New[k]-sum[vec[i][k]]);
	}
	ql=qr=n;ans=Query(1,0,n)+sum[n];
	printf("%d\n",n-ans);
}
### 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<bits/stdc++.h>
#define X first
#define Y second
#define N 200005
using namespace std;
vector<int>vec[N];
int T[N<<2],b[N],sum[N],New[N];
int ql,qr,n,i,k,ans,m,Zero;
void build(int x,int l,int r){
	T[x]=-1e9;if (l==r) return;
  int mid=(l+r)>>1;
  build(x<<1,l,mid);
  build(x<<1|1,mid+1,r);
}
int Query(int x,int l,int r){
	if (ql<=l&&r<=qr) return T[x];
	int mid=(l+r)>>1,ret=-1e9;
	if (ql<=mid) ret=max(ret,Query(x<<1,l,mid));
	if (qr>mid)  ret=max(ret,Query(x<<1|1,mid+1,r));
	return ret;
}
void Update(int x,int l,int r,int pos,int val){
	if (l==r) {T[x]=max(T[x],val);return;}
	int mid=(l+r)>>1;
	if (pos<=mid) Update(x<<1,l,mid,pos,val);
	         else Update(x<<1|1,mid+1,r,pos,val);
	T[x]=max(T[x<<1],T[x<<1|1]);
}
int main(){
	scanf("%d",&n);
	for (i=1;i<=n;i++) 
	  scanf("%d",&b[i]),sum[i]=sum[i-1]+b[i];
	scanf("%d",&m);
	for (i=1;i<=m;i++)
	  scanf("%d%d",&ql,&qr),vec[ql].push_back(qr);
	build(1,0,n); 
	Update(1,0,n,0,0-sum[0]);
	for (i=1;i<=n;i++){
		ql=qr=i-1;Zero=Query(1,0,n)+sum[i-1]+(b[i]==0);
		for (k=0;k<vec[i].size();k++){
			ql=i-1;qr=vec[i][k];
			New[k]=sum[vec[i][k]]+Query(1,0,n);
		}
		Update(1,0,n,i,Zero-sum[i]);
		for (k=0;k<vec[i].size();k++)
		  Update(1,0,n,vec[i][k],New[k]-sum[vec[i][k]]);
	}
	ql=qr=n;ans=Query(1,0,n)+sum[n];
	printf("%d\n",n-ans);
}
```