//ΔARC085D
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 222222;
const int inf = 1e9;
int n,q,b[N],f[N],g[N];
pair<int,int> p[N];
int T[N<<2];
void build(int k=1,int l=0,int r=n){
	if(l==r){
		T[k]=inf;
		return;
	}
	int h=l+r>>1;
	build(k<<1,l,h);
	build(k<<1|1,h+1,r);
	T[k]=min(T[k<<1],T[k<<1|1]);
}
void modify(int x,int y,int k=1,int l=0,int r=n){
	if(l==r){
		T[k]=y;
		return;
	}
	int h=l+r>>1;
	if(x<=h)
		modify(x,y,k<<1,l,h);
	else
		modify(x,y,k<<1|1,h+1,r);
	T[k]=min(T[k<<1],T[k<<1|1]);
}
int query(int L,int R,int k=1,int l=0,int r=n){
	if(L<=l&&r<=R)
		return T[k];
	int h=l+r>>1,res=inf;
	if(L<=h)
		res=min(res,query(L,R,k<<1,l,h));
	if(h<R)
		res=min(res,query(L,R,k<<1|1,h+1,r));
	return res;
}
int main()
{
	int i,x,y;
	scanf("%d",&n);
	for(x=1;x<=n;x=x+1)
		f[x]=inf;
	for(i=1;i<=n;i=i+1){
		scanf("%d",b+i);
		if(!b[i])
			b[i]=-1,f[0]++;
	}
	scanf("%d",&q);
	for(i=1;i<=q;i=i+1){
		scanf("%d%d",&x,&y);
		p[i]=make_pair(x,-y);
	}
	sort(p+1,p+q+1);
	build();
	modify(0,f[0]);
	i=1;
	for(x=1;x<=n;x=x+1){
		while(p[i].first==x){
			y=-p[i++].second;
			f[y]=min(f[y],query(x-1,y));
			modify(y,f[y]);
		}
		f[x]=min(f[x],f[x-1]+b[x]);
		modify(x,f[x]);
	}
	printf("%d",f[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
//ΔARC085D
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 222222;
const int inf = 1e9;
int n,q,b[N],f[N],g[N];
pair<int,int> p[N];
int T[N<<2];
void build(int k=1,int l=0,int r=n){
	if(l==r){
		T[k]=inf;
		return;
	}
	int h=l+r>>1;
	build(k<<1,l,h);
	build(k<<1|1,h+1,r);
	T[k]=min(T[k<<1],T[k<<1|1]);
}
void modify(int x,int y,int k=1,int l=0,int r=n){
	if(l==r){
		T[k]=y;
		return;
	}
	int h=l+r>>1;
	if(x<=h)
		modify(x,y,k<<1,l,h);
	else
		modify(x,y,k<<1|1,h+1,r);
	T[k]=min(T[k<<1],T[k<<1|1]);
}
int query(int L,int R,int k=1,int l=0,int r=n){
	if(L<=l&&r<=R)
		return T[k];
	int h=l+r>>1,res=inf;
	if(L<=h)
		res=min(res,query(L,R,k<<1,l,h));
	if(h<R)
		res=min(res,query(L,R,k<<1|1,h+1,r));
	return res;
}
int main()
{
	int i,x,y;
	scanf("%d",&n);
	for(x=1;x<=n;x=x+1)
		f[x]=inf;
	for(i=1;i<=n;i=i+1){
		scanf("%d",b+i);
		if(!b[i])
			b[i]=-1,f[0]++;
	}
	scanf("%d",&q);
	for(i=1;i<=q;i=i+1){
		scanf("%d%d",&x,&y);
		p[i]=make_pair(x,-y);
	}
	sort(p+1,p+q+1);
	build();
	modify(0,f[0]);
	i=1;
	for(x=1;x<=n;x=x+1){
		while(p[i].first==x){
			y=-p[i++].second;
			f[y]=min(f[y],query(x-1,y));
			modify(y,f[y]);
		}
		f[x]=min(f[x],f[x-1]+b[x]);
		modify(x,f[x]);
	}
	printf("%d",f[n]);
	return 0;
}
```