#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1) 
#define mid ((l+r)/2)
#define ll long long
#define pb(x) push_back(x)
#define pr(x) cerr<<#x<<"="<<(x)<<endl
#define pri(x,lo) {cerr<<#x<<"={";for (int ol=0;ol<=lo;ol++)cerr<<x[ol]<<",";cerr<<"}"<<endl;}
#define N 1001000
#define inf 100000000
int minn[N],b[N],n,ans,i,j,m,o,p;
vector<int> s[N];
struct node{int l,r;}q[N];
bool cmp(node a,node b){if (a.l==b.l) return a.r>b.r;return a.l<b.l;}
void build(int rt,int l,int r)
{
	minn[rt]=inf;
	if (l==r) return ;
	build(ls,l,mid),build(rs,mid+1,r);
}
void modify(int rt,int l,int r,int x,int y)
{
	if (l==r) {minn[rt]=min(minn[rt],y);return;}
	if (x<=mid) modify(ls,l,mid,x,y);
	else modify(rs,mid+1,r,x,y);
	minn[rt]=min(minn[ls],minn[rs]);
}
int query(int rt,int l,int r,int x,int y)
{
	int ret=inf;
	if (x<=l&&r<=y) return minn[rt];
	if (x<=mid) ret=min(ret,query(ls,l,mid,x,y));
	if (y>=mid+1) ret=min(ret,query(rs,mid+1,r,x,y));
	return ret;
}
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&b[i]),ans+=1-b[i];
	scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d %d",&o,&p),q[i].l=o,q[i].r=p;
	sort(q+1,q+1+m,cmp);
	for (i=1;i<=m;i++) s[q[i].l].pb(q[i].r);
	build(1,0,n);
	modify(1,0,n,0,0);
	for (i=1;i<=n;i++) 
	{
		for (j=0;j<s[i].size();j++)
		{
			int tmp=query(1,0,n,i-1,s[i][j]);
			modify(1,0,n,s[i][j],tmp);
		}
		modify(1,0,n,i,query(1,0,n,i-1,i-1)+(b[i]?1:-1));
	}
	printf("%d\n",ans+query(1,0,n,n,n));
	return 0;
}
### Prompt

In CPP, your task is to solve 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1) 
#define mid ((l+r)/2)
#define ll long long
#define pb(x) push_back(x)
#define pr(x) cerr<<#x<<"="<<(x)<<endl
#define pri(x,lo) {cerr<<#x<<"={";for (int ol=0;ol<=lo;ol++)cerr<<x[ol]<<",";cerr<<"}"<<endl;}
#define N 1001000
#define inf 100000000
int minn[N],b[N],n,ans,i,j,m,o,p;
vector<int> s[N];
struct node{int l,r;}q[N];
bool cmp(node a,node b){if (a.l==b.l) return a.r>b.r;return a.l<b.l;}
void build(int rt,int l,int r)
{
	minn[rt]=inf;
	if (l==r) return ;
	build(ls,l,mid),build(rs,mid+1,r);
}
void modify(int rt,int l,int r,int x,int y)
{
	if (l==r) {minn[rt]=min(minn[rt],y);return;}
	if (x<=mid) modify(ls,l,mid,x,y);
	else modify(rs,mid+1,r,x,y);
	minn[rt]=min(minn[ls],minn[rs]);
}
int query(int rt,int l,int r,int x,int y)
{
	int ret=inf;
	if (x<=l&&r<=y) return minn[rt];
	if (x<=mid) ret=min(ret,query(ls,l,mid,x,y));
	if (y>=mid+1) ret=min(ret,query(rs,mid+1,r,x,y));
	return ret;
}
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&b[i]),ans+=1-b[i];
	scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d %d",&o,&p),q[i].l=o,q[i].r=p;
	sort(q+1,q+1+m,cmp);
	for (i=1;i<=m;i++) s[q[i].l].pb(q[i].r);
	build(1,0,n);
	modify(1,0,n,0,0);
	for (i=1;i<=n;i++) 
	{
		for (j=0;j<s[i].size();j++)
		{
			int tmp=query(1,0,n,i-1,s[i][j]);
			modify(1,0,n,s[i][j],tmp);
		}
		modify(1,0,n,i,query(1,0,n,i-1,i-1)+(b[i]?1:-1));
	}
	printf("%d\n",ans+query(1,0,n,n,n));
	return 0;
}
```