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

const int maxn=2e5+10;
int n,q,a[maxn],dp[maxn];
int st[maxn<<2];
vector<int> g[maxn];

inline int read(){
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	x=c-48;
	while((c=getchar())>='0'&&c<='9')x=x*10+c-48;
	return x;
}
inline void chkmin(int&a,int b){
	if(a>b)a=b;
}
inline int min_(int a,int b){
	return a<=b?a:b;
}
inline void push_up(int rt){
	st[rt]=min_(st[rt<<1],st[rt<<1|1]);
}
void build(int rt,int l,int r){
	if(l==r){
		st[rt]=dp[l]-a[l];
		return;
	}
	int mid=l+r>>1;
	build(rt<<1,l,mid);
	build(rt<<1|1,mid+1,r);
	push_up(rt);
}
void modify(int rt,int l,int r,int pos){
	if(l==r){
		st[rt]=dp[pos]-a[pos];
		return;
	}
	int mid=l+r>>1;
	if(pos<=mid)
		modify(rt<<1,l,mid,pos);
	else
		modify(rt<<1|1,mid+1,r,pos);
	push_up(rt);
}
int query(int rt,int l,int r,int x,int y){
	if(l==x&&r==y)
		return st[rt];
	int mid=l+r>>1;
	if(y<=mid)
		return query(rt<<1,l,mid,x,y);
	else if(x>mid)
		return query(rt<<1|1,mid+1,r,x,y);
	else
		return min_(query(rt<<1,l,mid,x,mid),query(rt<<1|1,mid+1,r,mid+1,y));
}

int main(){
	n=read();
	for(int i=1;i<=n;++i)
		a[i]=a[i-1]+(read()^1);
	q=read();
	for(int i=1,l,r;i<=q;++i){
		l=read();r=read();
		g[l-1].push_back(r);
	}
	for(int i=1;i<=n;++i)
		dp[i]=2e9;
	build(1,0,n);
	for(int i=0;i<n;++i){
		for(int j=0;j<g[i].size();++j){
			chkmin(dp[g[i][j]],query(1,0,n,i,g[i][j]-1)+a[g[i][j]]);
			modify(1,0,n,g[i][j]);
		}
		chkmin(dp[i+1],dp[i]+(a[i+1]-a[i]^1));
		modify(1,0,n,i+1);
	}
	printf("%d\n",dp[n]);
	return 0;
}
### Prompt

Develop a solution in cpp 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 maxn=2e5+10;
int n,q,a[maxn],dp[maxn];
int st[maxn<<2];
vector<int> g[maxn];

inline int read(){
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	x=c-48;
	while((c=getchar())>='0'&&c<='9')x=x*10+c-48;
	return x;
}
inline void chkmin(int&a,int b){
	if(a>b)a=b;
}
inline int min_(int a,int b){
	return a<=b?a:b;
}
inline void push_up(int rt){
	st[rt]=min_(st[rt<<1],st[rt<<1|1]);
}
void build(int rt,int l,int r){
	if(l==r){
		st[rt]=dp[l]-a[l];
		return;
	}
	int mid=l+r>>1;
	build(rt<<1,l,mid);
	build(rt<<1|1,mid+1,r);
	push_up(rt);
}
void modify(int rt,int l,int r,int pos){
	if(l==r){
		st[rt]=dp[pos]-a[pos];
		return;
	}
	int mid=l+r>>1;
	if(pos<=mid)
		modify(rt<<1,l,mid,pos);
	else
		modify(rt<<1|1,mid+1,r,pos);
	push_up(rt);
}
int query(int rt,int l,int r,int x,int y){
	if(l==x&&r==y)
		return st[rt];
	int mid=l+r>>1;
	if(y<=mid)
		return query(rt<<1,l,mid,x,y);
	else if(x>mid)
		return query(rt<<1|1,mid+1,r,x,y);
	else
		return min_(query(rt<<1,l,mid,x,mid),query(rt<<1|1,mid+1,r,mid+1,y));
}

int main(){
	n=read();
	for(int i=1;i<=n;++i)
		a[i]=a[i-1]+(read()^1);
	q=read();
	for(int i=1,l,r;i<=q;++i){
		l=read();r=read();
		g[l-1].push_back(r);
	}
	for(int i=1;i<=n;++i)
		dp[i]=2e9;
	build(1,0,n);
	for(int i=0;i<n;++i){
		for(int j=0;j<g[i].size();++j){
			chkmin(dp[g[i][j]],query(1,0,n,i,g[i][j]-1)+a[g[i][j]]);
			modify(1,0,n,g[i][j]);
		}
		chkmin(dp[i+1],dp[i]+(a[i+1]-a[i]^1));
		modify(1,0,n,i+1);
	}
	printf("%d\n",dp[n]);
	return 0;
}
```