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

Solution (in CPP):

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

inline int read()
{
	int sum=0,ff=1; char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-') ff=-1;
		ch=getchar();
	}
	while(isdigit(ch))
		sum=sum*10+(ch^48),ch=getchar();
	return sum*ff;
}

const int N=2e5+5;

int n,m,suf[N][2],laz[N<<2],tr[N<<2];
int ans,a[N],f[N];
vector<int> po[N];

struct nood
{
	int l,r;
	inline bool friend operator < (const nood&c,const nood &d)
	{
		return c.l<d.l;
	}
};
nood c[N];

inline int min(int x,int y) 
{
	return (x>y)?y:x;
}

inline void build(int rt,int l,int r)
{
	tr[rt]=1e9;
	if(l==r) return;
	int mid=(l+r)/2;
	build(rt<<1,l,mid); 
	build(rt<<1|1,mid+1,r);
	tr[rt]=min(tr[rt<<1],tr[rt<<1|1]);
}

inline void modify(int rt,int l,int r,int pos,int val)
{
	if(l==r)
	{
		tr[rt]=min(tr[rt],val);
		return;
	}
	int mid=(l+r)/2;
	if(pos<=mid) modify(rt<<1,l,mid,pos,val);
	else modify(rt<<1|1,mid+1,r,pos,val);
	tr[rt]=min(tr[rt<<1],tr[rt<<1|1]);
}

inline int query(int rt,int l,int r,int ll,int rr)
{
	if(ll<=l&&r<=rr) return tr[rt];
	int mid=(l+r)/2;
	int ret=1e9;
	if(ll<=mid) ret=min(ret,query(rt<<1,l,mid,ll,rr));
	if(rr>mid) ret=min(ret,query(rt<<1|1,mid+1,r,ll,rr));
	return ret;
}

signed main()
{
	n=read();
	for ( int i=1;i<=n;i++ ) 
	{
		int x=read();
		a[i]=x;
		suf[i][0]=suf[i-1][0]+(x==0);
		suf[i][1]=suf[i-1][1]+(x==1);
	}
	memset(f,127/3,sizeof(f));
	m=read();
	for ( int i=1;i<=m;i++ ) c[i]=(nood){read(),read()};
	sort(c+1,c+m+1);
	for ( int i=1;i<=m;i++ ) po[c[i].l].push_back(c[i].r);
	f[0]=0;
	build(1,1,n);
	for ( int i=1;i<=n;i++ ) 
	{
		for ( int j=0;j<po[i].size();j++ )
		{
			int r=po[i][j];
			int mi=f[i-1];
			mi=min(mi,query(1,1,n,max(i-1,1),r));
			if(mi<f[r])
			{
				f[r]=mi;
				modify(1,1,n,r,mi);
			}
		}
		f[i]=min(f[i],f[i-1]+((a[i]==1)?1:-1));
	}
	printf("%d\n",f[n]+suf[n][0]);
	return 0;
}